Equations of circle Questions in English

(C) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the given points $(-4, 3)$ and $(12, -1)$:
$(x + 4)(x - 12) + (y - 3)(y + 1) = 0$
$x^2 - 12x + 4x - 48 + y^2 + y - 3y - 3 = 0$
$x^2 + y^2 - 8x - 2y - 51 = 0$.
The length of the intercept on the $y$-axis is given by $2 \sqrt{f^2 - c}$,where the equation is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Here,$2f = -2 \implies f = -1$ and $c = -51$.
Intercept length $= 2 \sqrt{(-1)^2 - (-51)} = 2 \sqrt{1 + 51} = 2 \sqrt{52} = 2 \sqrt{4 \times 13} = 4 \sqrt{13}$.

(B) Since the circle touches the $X$-axis at the origin $(0, 0)$,its center must lie on the $Y$-axis at $(0, a)$ or $(0, -a)$.
Thus,the radius is $a$.
The equation of the circle is $(x - 0)^2 + (y \mp a)^2 = a^2$.
Expanding this,we get $x^2 + y^2 \mp 2ay + a^2 = a^2$.
Therefore,the equation is $x^2 + y^2 \pm 2ay = 0$.

(A) The center of the circle is $(h, k) = (2, 1)$.
Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the center.
Thus,$r = |1| = 1$.
The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - 2)^2 + (y - 1)^2 = 1^2$.
Expanding this,we have $(x^2 - 4x + 4) + (y^2 - 2y + 1) = 1$.
Simplifying,$x^2 + y^2 - 4x - 2y + 4 = 0$.

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the circle passes through the origin $(0, 0)$,we have $c = 0$.
The center of the circle is $(-g, -f)$.
Since the center lies on the $y$-axis,the $x$-coordinate of the center must be zero,so $-g = 0$,which implies $g = 0$.
The equation of the circle becomes $x^2 + y^2 + 2fy = 0$.
The radius of the circle is given by $\sqrt{g^2 + f^2 - c} = \sqrt{0^2 + f^2 - 0} = |f|$.
Given that the radius is $3$,we have $|f| = 3$,which means $f = 3$ or $f = -3$.
Substituting $f = 3$,we get $x^2 + y^2 + 6y = 0$.
Substituting $f = -3$,we get $x^2 + y^2 - 6y = 0$.
Comparing with the given options,$x^2 + y^2 + 6y = 0$ is the correct choice.

(A) The center of the circle is the intersection of the two diameters $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$.
Multiplying the second equation by $3$,we get $9x - 3y - 12 = 0$.
Adding this to the first equation: $(2x + 3y + 1) + (9x - 3y - 12) = 0 \implies 11x - 11 = 0 \implies x = 1$.
Substituting $x = 1$ into $3x - y - 4 = 0$,we get $3(1) - y - 4 = 0 \implies y = -1$.
So,the center $(h, k) = (1, -1)$.
The circumference is $2\pi r = 10\pi$,which gives $r = 5$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y + 1)^2 = 5^2$.
$x^2 - 2x + 1 + y^2 + 2y + 1 = 25$.
$x^2 + y^2 - 2x + 2y - 23 = 0$.

(A) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
The radius $R$ of this circle is given by $R = \sqrt{g^2 + f^2 - c}$.
For the circle to be a point circle,its radius must be zero,i.e.,$R = 0$.
Therefore,$\sqrt{g^2 + f^2 - c} = 0$.
Squaring both sides,we get $g^2 + f^2 - c = 0$,which implies $g^2 + f^2 = c$.

(A) Two circles are concentric if they have the same center. The given circle is $x^2 + y^2 - 2x + 4y + 20 = 0$.
The equation of any circle concentric to this circle is of the form $x^2 + y^2 - 2x + 4y + c = 0$.
Since this circle passes through the point $(4, -2)$,we substitute $x = 4$ and $y = -2$ into the equation:
$(4)^2 + (-2)^2 - 2(4) + 4(-2) + c = 0$
$16 + 4 - 8 - 8 + c = 0$
$20 - 16 + c = 0$
$4 + c = 0$
$c = -4$

(A) The given circle equation is $x^2 + y^2 - 2gx + f^2 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,the center of this circle is $(g, 0)$.
The required circle has center $(a, b)$ and passes through the point $(g, 0)$.
The radius $r$ is the distance between the center $(a, b)$ and the point $(g, 0)$.
Using the distance formula,$r = \sqrt{(a - g)^2 + (b - 0)^2} = \sqrt{(a - g)^2 + b^2}$.

(C) To find the center $(h, k)$,solve the system of equations:
$2x - 3y = -4$ $(i)$
$3x + 4y = 5$ (ii)
Multiplying $(i)$ by $4$ and (ii) by $3$: $8x - 12y = -16$ and $9x + 12y = 15$.
Adding them gives $17x = -1$,so $x = -\frac{1}{17}$.
Substituting $x$ into $(i)$: $2(-\frac{1}{17}) - 3y = -4 \implies -3y = -4 + \frac{2}{17} = -\frac{66}{17} \implies y = \frac{22}{17}$.
Center is $(h, k) = (-\frac{1}{17}, \frac{22}{17})$.
Since the circle passes through the origin $(0, 0)$,the radius $r$ is the distance from the center to the origin:
$r^{2} = h^{2} + k^{2} = (-\frac{1}{17})^{2} + (\frac{22}{17})^{2} = \frac{1 + 484}{289} = \frac{485}{289}$.
The equation of the circle is $(x - h)^{2} + (y - k)^{2} = r^{2}$:
$(x + \frac{1}{17})^{2} + (y - \frac{22}{17})^{2} = \frac{485}{289}$
$x^{2} + \frac{2x}{17} + \frac{1}{289} + y^{2} - \frac{44y}{17} + \frac{484}{289} = \frac{485}{289}$
$x^{2} + y^{2} + \frac{2x}{17} - \frac{44y}{17} + \frac{485}{289} = \frac{485}{289}$
$x^{2} + y^{2} + \frac{2x}{17} - \frac{44y}{17} = 0$
Multiplying by $17$: $17(x^{2} + y^{2}) + 2x - 44y = 0$.

(C) The given circle passes through the origin,so the constant term must be zero: $p - 3 = 0 \Rightarrow p = 3$.
Substituting $p = 3$ into the equation,we get $x^{2} + y^{2} + 9x - 3y = 0$.
The radius $r_1$ of this circle is $\sqrt{(\frac{9}{2})^2 + (-\frac{3}{2})^2} = \sqrt{\frac{81}{4} + \frac{9}{4}} = \sqrt{\frac{90}{4}} = \frac{3\sqrt{10}}{2}$.
The required circle passes through the origin,so its equation is of the form $x^{2} + y^{2} + 2gx + 2fy = 0$.
Its radius $r_2$ is $\sqrt{g^2 + f^2}$.
Given $r_2 = 2r_1$,so $\sqrt{g^2 + f^2} = 2 \times \frac{3\sqrt{10}}{2} = 3\sqrt{10}$.
Thus,$g^2 + f^2 = 90$.
Since the circle passes through the origin,the tangent at the origin for the first circle is $9x - 3y = 0$,which implies the slope is $3$. For the new circle,the tangent is $gx + fy = 0$,implying the slope is $-g/f$. Assuming the circles are tangent at the origin,$-g/f = 3 \Rightarrow g = -3f$.
Substituting into $g^2 + f^2 = 90$: $(-3f)^2 + f^2 = 90$ $\Rightarrow 10f^2 = 90$ $\Rightarrow f^2 = 9$ $\Rightarrow f = \pm 3$.
If $f = -3$,then $g = 9$. The equation is $x^{2} + y^{2} + 18x - 6y = 0$.

(A) The vertices of the square are formed by the intersection of the lines $x=2, x=3, y=1, y=2$.
These vertices are $A(2, 1), B(3, 1), C(3, 2),$ and $D(2, 2)$.
The diagonals of the square are $AC$ and $BD$.
If we take the diagonal $AC$ as the diameter,the endpoints are $(2, 1)$ and $(3, 2)$.
The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the values for $AC$: $(x - 2)(x - 3) + (y - 1)(y - 2) = 0$.
Expanding this: $x^2 - 5x + 6 + y^2 - 3y + 2 = 0$.
Thus,$x^2 + y^2 - 5x - 3y + 8 = 0$.
Similarly,if we take the diagonal $BD$ as the diameter,the endpoints are $(2, 2)$ and $(3, 1)$.
The equation is $(x - 2)(x - 3) + (y - 2)(y - 1) = 0$,which results in the same equation: $x^2 + y^2 - 5x - 3y + 8 = 0$.

(A) Let $A = (x_{1}, y_{1})$ and $B = (x_{2}, y_{2})$.
From the given equations,we have:
$x_{1} + x_{2} = -2a$ and $x_{1}x_{2} = -b^{2}$
$y_{1} + y_{2} = -2p$ and $y_{1}y_{2} = -q^{2}$
The equation of the circle with $AB$ as diameter is $(x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0$.
Expanding this,we get $x^{2} - (x_{1} + x_{2})x + x_{1}x_{2} + y^{2} - (y_{1} + y_{2})y + y_{1}y_{2} = 0$.
Substituting the values,we get $x^{2} + 2ax - b^{2} + y^{2} + 2py - q^{2} = 0$,which is $x^{2} + y^{2} + 2ax + 2py - (b^{2} + q^{2}) = 0$.
The standard form of a circle is $x^{2} + y^{2} + 2gx + 2fy + c = 0$,where the radius is $\sqrt{g^{2} + f^{2} - c}$.
Here,$g = a$,$f = p$,and $c = -(b^{2} + q^{2})$.
Therefore,the radius is $\sqrt{a^{2} + p^{2} - (-(b^{2} + q^{2}))} = \sqrt{a^{2} + p^{2} + b^{2} + q^{2}}$.

(A) The center of the circle is the point of intersection of the two diameters.
Solving the system of equations:
$2x - 3y = 5$ $(i)$
$3x - 4y = 7$ $(ii)$
Multiplying $(i)$ by $3$ and $(ii)$ by $2$:
$6x - 9y = 15$
$6x - 8y = 14$
Subtracting the equations: $-y = 1 \Rightarrow y = -1$.
Substituting $y = -1$ into $(i)$:
$2x - 3(-1) = 5$ $\Rightarrow 2x + 3 = 5$ $\Rightarrow 2x = 2$ $\Rightarrow x = 1$.
So,the center $(h, k)$ is $(1, -1)$ and the radius $r = 8$.
The equation of the circle is $(x - h)^{2} + (y - k)^{2} = r^{2}$.
$(x - 1)^{2} + (y - (-1))^{2} = 8^{2}$
$(x - 1)^{2} + (y + 1)^{2} = 64$
$x^{2} - 2x + 1 + y^{2} + 2y + 1 = 64$
$x^{2} + y^{2} - 2x + 2y + 2 = 64$
$x^{2} + y^{2} - 2x + 2y - 62 = 0$.

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $(i)$.
Since the circle passes through $(1, 0)$ and $(0, 1)$:
For $(1, 0)$: $1 + 2g + c = 0$ $(ii)$.
For $(0, 1)$: $1 + 2f + c = 0$ $(iii)$.
Subtracting $(iii)$ from $(ii)$,we get $2g - 2f = 0$,so $g = f$.
Substituting $f = g$ into $(ii)$,we get $c = -(1 + 2g)$.
The radius $R$ is given by $R = \sqrt{g^2 + f^2 - c} = \sqrt{g^2 + g^2 - (-(1 + 2g))} = \sqrt{2g^2 + 2g + 1}$.
To minimize $R$,we minimize $R^2 = 2g^2 + 2g + 1$.
Differentiating with respect to $g$: $\frac{d}{dg}(2g^2 + 2g + 1) = 4g + 2 = 0$,which gives $g = -\frac{1}{2}$.
Since $g = f$,we have $f = -\frac{1}{2}$.
Then $c = -(1 + 2(-\frac{1}{2})) = -(1 - 1) = 0$.
Substituting $g, f, c$ into $(i)$,the equation is $x^2 + y^2 - x - y = 0$.
Alternatively,the circle with minimum radius passing through two points has the segment joining those points as its diameter.
The equation is $(x - 1)(x - 0) + (y - 0)(y - 1) = 0$,which simplifies to $x^2 - x + y^2 - y = 0$ or $x^2 + y^2 - x - y = 0$.

(A) The given equation of the circle is $2x^2 + 2y^2 = 3x - 5y + 7$.
Dividing by $2$,we get $x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -\frac{3}{2} \implies g = -\frac{3}{4}$ and $2f = \frac{5}{2} \implies f = \frac{5}{4}$.
The center of the circle is $(-g, -f) = \left( \frac{3}{4}, -\frac{5}{4} \right)$.
The radius $r$ is given by $\sqrt{g^2 + f^2 - c} = \sqrt{\left( -\frac{3}{4} \right)^2 + \left( \frac{5}{4} \right)^2 - \left( -\frac{7}{2} \right)}$.
$r = \sqrt{\frac{9}{16} + \frac{25}{16} + \frac{7}{2}} = \sqrt{\frac{9 + 25 + 56}{16}} = \sqrt{\frac{90}{16}} = \frac{3\sqrt{10}}{4}$.

(C) The initial center of the circle is $C(5, 5)$ as it touches both axes in the first quadrant with radius $r = 5$.
When the circle rolls one complete revolution along the $x-$axis,the distance covered by the center is equal to the circumference of the circle,which is $2\pi r = 2\pi(5) = 10\pi$.
The new center $D$ will have coordinates $(5 + 10\pi, 5)$ while the radius remains $5$.
The equation of the circle in the new position is $(x - (5 + 10\pi))^{2} + (y - 5)^{2} = 5^{2}$,which simplifies to $(x - 5 - 10\pi)^{2} + (y - 5)^{2} = 25$.

(C) Four points $(x_1, y_1), (x_2, y_2), (x_3, y_3),$ and $(x_4, y_4)$ are concyclic if the cross-ratio of their complex numbers is real,or equivalently,if the determinant of the matrix formed by their coordinates is zero.
Alternatively,for four points to be concyclic,the power of the point $(0, t)$ with respect to the circle passing through $(2, 3), (0, 2),$ and $(4, 5)$ must be zero.
Let the circle equation be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Substituting $(2, 3): 4 + 9 + 4g + 6f + c = 0 \Rightarrow 4g + 6f + c = -13$.
Substituting $(0, 2): 0 + 4 + 0g + 4f + c = 0 \Rightarrow 4f + c = -4$.
Substituting $(4, 5): 16 + 25 + 8g + 10f + c = 0 \Rightarrow 8g + 10f + c = -41$.
Subtracting the first from the third: $4g + 4f = -28 \Rightarrow g + f = -7$.
From $4f + c = -4$,we have $c = -4 - 4f$.
Substitute into the first: $4g + 6f - 4 - 4f = -13 \Rightarrow 4g + 2f = -9$.
Solving $g+f = -7$ and $4g+2f = -9$: $2g + 2f = -14$ and $4g + 2f = -9 \Rightarrow 2g = 5 \Rightarrow g = 2.5, f = -9.5$.
Then $c = -4 - 4(-9.5) = -4 + 38 = 34$.
The circle is $x^2 + y^2 + 5x - 19y + 34 = 0$.
For $(0, t)$ to lie on this circle: $0^2 + t^2 + 5(0) - 19(t) + 34 = 0 \Rightarrow t^2 - 19t + 34 = 0$.
Factoring: $(t - 2)(t - 17) = 0$. Thus $t = 2$ or $t = 17$. Since the points must be distinct,$t = 17$ is the valid solution.

(A) The given circle is $x^{2} + y^{2} - 8x + 6y - 5 = 0$.
Comparing this with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -4$ and $f = 3$.
The center of the circle is $(-g, -f) = (4, -3)$.
Since the required circle is concentric with the given circle,its center is also $(4, -3)$.
The circle passes through the point $(-2, -7)$.
The radius $r$ is the distance between $(4, -3)$ and $(-2, -7)$:
$r^{2} = (4 - (-2))^{2} + (-3 - (-7))^{2} = (6)^{2} + (4)^{2} = 36 + 16 = 52$.
The equation of the circle is $(x - 4)^{2} + (y + 3)^{2} = 52$.
Expanding this,we get $x^{2} - 8x + 16 + y^{2} + 6y + 9 = 52$,which simplifies to $x^{2} + y^{2} - 8x + 6y - 27 = 0$.

(C) The equation of a circle with center $(3, 1)$ is $(x - 3)^2 + (y - 1)^2 = r^2$.
Since the circle touches the line $8x - 15y + 25 = 0$,the radius $r$ is equal to the perpendicular distance from the center $(3, 1)$ to the line.
$r = \frac{|8(3) - 15(1) + 25|}{\sqrt{8^2 + (-15)^2}}$
$r = \frac{|24 - 15 + 25|}{\sqrt{64 + 225}}$
$r = \frac{|34|}{\sqrt{289}} = \frac{34}{17} = 2$.
Substituting $r = 2$ into the circle equation:
$(x - 3)^2 + (y - 1)^2 = 2^2$
$(x^2 - 6x + 9) + (y^2 - 2y + 1) = 4$
$x^2 + y^2 - 6x - 2y + 10 = 4$
$x^2 + y^2 - 6x - 2y + 6 = 0$.

(A) Since the circle touches both axes in the first quadrant,its center is $(r, r)$ and radius is $r$,where $r > 0$.
The equation of the circle is $(x - r)^2 + (y - r)^2 = r^2$,which simplifies to $x^2 + y^2 - 2rx - 2ry + r^2 = 0$.
The distance from the center $(r, r)$ to the line $4x + 3y - 6 = 0$ must be greater than the radius $r$ for the circle to lie below the line.
The perpendicular distance $d = \frac{|4r + 3r - 6|}{\sqrt{4^2 + 3^2}} = \frac{|7r - 6|}{5}$.
For the circle to be below the line,we require $d > r$ or specific conditions. However,if the circle is tangent to the line,$d = r$.
$\frac{|7r - 6|}{5} = r \implies |7r - 6| = 5r$.
Case $1$: $7r - 6 = 5r \implies 2r = 6 \implies r = 3$.
Case $2$: $7r - 6 = -5r \implies 12r = 6 \implies r = 0.5$.
For $r = 0.5$,the circle is $(x - 0.5)^2 + (y - 0.5)^2 = 0.25$,which is $x^2 + y^2 - x - y + 0.25 = 0$,or $4x^2 + 4y^2 - 4x - 4y + 1 = 0$.

(A) Let the points be $A(1, 0)$ and $B(0, 1)$.
For a circle passing through two points to have the smallest radius,the line segment joining these two points must be the diameter of the circle.
The midpoint of $AB$ is the center of the circle: $C = (\frac{1+0}{2}, \frac{0+1}{2}) = (\frac{1}{2}, \frac{1}{2})$.
The diameter is the distance $AB = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$.
The radius $r = \frac{\text{diameter}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
The equation of the circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the values: $(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = (\frac{1}{\sqrt{2}})^2$.
$x^2 - x + \frac{1}{4} + y^2 - y + \frac{1}{4} = \frac{1}{2}$.
$x^2 + y^2 - x - y + \frac{1}{2} = \frac{1}{2}$.
$x^2 + y^2 - x - y = 0$.

(C) The given equation of the circle is $x^2 + y^2 = 4$.
This is in the form $x^2 + y^2 = r^2$,where $r^2 = 4$,so the radius $r = 2$.
The parametric equations for a circle $x^2 + y^2 = r^2$ are given by $x = r \cos \alpha$ and $y = r \sin \alpha$.
Substituting $r = 2$,we get $x = 2 \cos \alpha$ and $y = 2 \sin \alpha$.
Thus,the coordinates of any point on the circle are $(2 \cos \alpha, 2 \sin \alpha)$.

(A) Since the circle touches both axes and lies in the fourth quadrant with radius $r = 1$,the center of the circle must be $(1, -1)$.
The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting $h = 1$,$k = -1$,and $r = 1$:
$(x - 1)^2 + (y - (-1))^2 = 1^2$
$(x - 1)^2 + (y + 1)^2 = 1$
Expanding the equation:
$(x^2 - 2x + 1) + (y^2 + 2y + 1) = 1$
$x^2 + y^2 - 2x + 2y + 2 = 1$
$x^2 + y^2 - 2x + 2y + 1 = 0$

(D) Let the center of the circle be $(h, k)$.
Since the circle touches the $y$-axis at $(0, 2)$,the radius $r = |h|$ and the $y$-coordinate of the center $k = 2$.
Thus,the equation of the circle is $(x - h)^2 + (y - 2)^2 = h^2$.
Since the circle passes through $(-1, 0)$,we substitute these coordinates into the equation:
$(-1 - h)^2 + (0 - 2)^2 = h^2$
$1 + 2h + h^2 + 4 = h^2$
$2h + 5 = 0$
$h = -5/2$.
So,the equation of the circle is $(x + 5/2)^2 + (y - 2)^2 = (-5/2)^2$.
$(x + 5/2)^2 + (y - 2)^2 = 25/4$.
Now,check the options by substituting the points into the equation:
For $(-4, 0)$:
$(-4 + 5/2)^2 + (0 - 2)^2 = (-8/2 + 5/2)^2 + (-2)^2 = (-3/2)^2 + 4 = 9/4 + 4 = 25/4$.
Since the point $(-4, 0)$ satisfies the equation,the circle passes through $(-4, 0)$.

(D) The circle touches the lines $x = 0$ and $x = 2c$. The distance between these two parallel lines is $2c$. Thus,the diameter of the circle is $2c$,and the radius $r = c$.
Since the circle touches $x = 0$ and $x = 2c$,the $x$-coordinate of the center must be $h = c$.
Since the circle also touches $y = 0$,the $y$-coordinate of the center must be $k = c$ or $k = -c$.
Thus,the center of the circle is $(c, c)$ or $(c, -c)$.
The equation of the circle is $(x - c)^2 + (y \mp c)^2 = c^2$.
Expanding this,we get $x^2 - 2cx + c^2 + y^2 \mp 2cy + c^2 = c^2$.
Simplifying,we get $x^2 + y^2 - 2cx \mp 2cy + c^2 = 0$.

(D) The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$,where the coefficients of $x^2$ and $y^2$ are equal and there is no $xy$ term.
Option $A$ has different coefficients for $x^2$ and $y^2$ ($1$ and $2$).
Option $B$ has different signs for $x^2$ and $y^2$.
Option $C$ contains an $xy$ term.
Option $D$ can be written as $3(x^2 + y^2) + 5x + 1 = 0$,which simplifies to $x^2 + y^2 + \frac{5}{3}x + \frac{1}{3} = 0$. This matches the form of a circle equation.

(C) For the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ to represent a circle,the following conditions must be satisfied:
$1$. The coefficient of $x^2$ must be equal to the coefficient of $y^2$ $(a = b)$.
$2$. The coefficient of $xy$ must be zero $(h = 0)$.
Given the equation $px^2 + (2 - q)xy + 3y^2 - 6qx + 30y + 6q = 0$:
Comparing the coefficients:
Coefficient of $x^2$ is $p$ and coefficient of $y^2$ is $3$.
Setting $p = 3$.
Coefficient of $xy$ is $(2 - q)$.
Setting $2 - q = 0$,we get $q = 2$.
Therefore,the values are $p = 3$ and $q = 2$.

(A) Let the center of the circle be $(h, k)$. Since the circle touches the $x$-axis and has a radius of $5$ in the first quadrant,the center must be $(h, 5)$ where $h > 0$.
The circle also touches the line $3x - 4y = 0$. The perpendicular distance from the center $(h, 5)$ to this line must be equal to the radius $5$.
$\Rightarrow \left|\frac{3h - 4(5)}{\sqrt{3^2 + (-4)^2}}\right| = 5$
$\Rightarrow \left|\frac{3h - 20}{5}\right| = 5$
$\Rightarrow |3h - 20| = 25$
This gives two cases: $3h - 20 = 25$ or $3h - 20 = -25$.
Case $1$: $3h = 45 \Rightarrow h = 15$.
Case $2$: $3h = -5 \Rightarrow h = -5/3$. Since the center is in the first quadrant,$h > 0$,so we take $h = 15$.
The equation of the circle is $(x - 15)^2 + (y - 5)^2 = 5^2$.
$x^2 - 30x + 225 + y^2 - 10y + 25 = 25$.
$x^2 + y^2 - 30x - 10y + 225 = 0$.

(C) The equation of a circle touching the $x$-axis at $(h, 0)$ is $(x-h)^2 + (y-k)^2 = k^2$,where $(h, k)$ is the center.
Given the circle touches the $x$-axis at $(3, 0)$,the equation is $(x-3)^2 + (y-k)^2 = k^2$.
Expanding this,we get $(x-3)^2 + y^2 - 2ky + k^2 = k^2$,which simplifies to $(x-3)^2 + y^2 - 2ky = 0$.
Since the circle passes through $(1, -2)$,we substitute these coordinates into the equation:
$(1-3)^2 + (-2)^2 - 2k(-2) = 0$
$(-2)^2 + 4 + 4k = 0$
$4 + 4 + 4k = 0$
$8 + 4k = 0$
$4k = -8 \Rightarrow k = -2$.
Substituting $k = -2$ back into the equation: $(x-3)^2 + y^2 - 2(-2)y = 0 \Rightarrow (x-3)^2 + y^2 + 4y = 0$.
Expanding further: $x^2 - 6x + 9 + y^2 + 4y = 0 \Rightarrow x^2 + y^2 - 6x + 4y + 9 = 0$.
Checking the options,for $(5, -2)$:
$(5)^2 + (-2)^2 - 6(5) + 4(-2) + 9 = 25 + 4 - 30 - 8 + 9 = 38 - 38 = 0$.
Thus,the circle passes through $(5, -2)$.

(A) The lines are $x = 0$ (y-axis),$y = 0$ (x-axis),and $\frac{x}{a} - \frac{y}{b} = 1$.
The vertices of the triangle are the intersection points of these lines:
$1$. Intersection of $x=0$ and $y=0$ is $O(0, 0)$.
$2$. Intersection of $x=0$ and $\frac{x}{a} - \frac{y}{b} = 1$ is $(0, -b)$.
$3$. Intersection of $y=0$ and $\frac{x}{a} - \frac{y}{b} = 1$ is $(a, 0)$.
Since the triangle is a right-angled triangle with vertices at $(0,0)$,$(a,0)$,and $(0,-b)$,the hypotenuse is the line segment connecting $(a,0)$ and $(0,-b)$.
The circumcircle of a right-angled triangle has the hypotenuse as its diameter.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting $(a, 0)$ and $(0, -b)$:
$(x - a)(x - 0) + (y - 0)(y - (-b)) = 0$
$x(x - a) + y(y + b) = 0$
$x^2 - ax + y^2 + by = 0$
$x^2 + y^2 - ax + by = 0$.

(B) The given equation of the circle is $(x - 1)^2 + (y - 1)^2 = 4$.
This is in the form $(x - h)^2 + (y - k)^2 = r^2$,where the center $(h, k) = (1, 1)$ and the radius $r = \sqrt{4} = 2$.
The parametric coordinates of a point on a circle with center $(h, k)$ and radius $r$ are given by $(h + r \cos \alpha, k + r \sin \alpha)$.
Substituting the values,we get $(1 + 2 \cos \alpha, 1 + 2 \sin \alpha)$.
Thus,the correct option is $B$.

(D) The given equation of the circle is $x^2 + y^2 - 4x - 6y + 4 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -4 \implies g = -2$ and $2f = -6 \implies f = -3$,with $c = 4$.
The radius $r$ of the circle is given by the formula $r = \sqrt{g^2 + f^2 - c}$.
Substituting the values: $r = \sqrt{(-2)^2 + (-3)^2 - 4} = \sqrt{4 + 9 - 4} = \sqrt{9} = 3$.
The diameter $d$ is $2 \times r = 2 \times 3 = 6$.

(A) Step-$1$: Find the radius $r$ of the circle.
The radius is the distance between the center $(2, -1)$ and the point $(3, 6)$ on the circle.
Using the distance formula $r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$r = \sqrt{(3 - 2)^2 + (6 - (-1))^2}$
$r = \sqrt{(1)^2 + (7)^2} = \sqrt{1 + 49} = \sqrt{50}$.
Step-$2$: Use the standard equation of a circle $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the center.
$(x - 2)^2 + (y - (-1))^2 = (\sqrt{50})^2$
$(x - 2)^2 + (y + 1)^2 = 50$
$x^2 - 4x + 4 + y^2 + 2y + 1 = 50$
$x^2 + y^2 - 4x + 2y + 5 = 50$
$x^2 + y^2 - 4x + 2y - 45 = 0$.

(B) The circle touches the $x$-axis at the point $(-3, 0)$.
Since the radius of the circle is $r = 4$,the center of the circle must be at $(-3, 4)$ or $(-3, -4)$.
Case $1$: Center is $(-3, 4)$ and radius $r = 4$.
The equation is $(x - (-3))^2 + (y - 4)^2 = 4^2$.
$(x + 3)^2 + (y - 4)^2 = 16$.
$x^2 + 6x + 9 + y^2 - 8y + 16 = 16$.
$x^2 + y^2 + 6x - 8y + 9 = 0$.
Case $2$: Center is $(-3, -4)$ and radius $r = 4$.
The equation is $(x - (-3))^2 + (y - (-4))^2 = 4^2$.
$(x + 3)^2 + (y + 4)^2 = 16$.
$x^2 + 6x + 9 + y^2 + 8y + 16 = 16$.
$x^2 + y^2 + 6x + 8y + 9 = 0$.
Combining both cases,the equation is $x^2 + y^2 + 6x \pm 8y + 9 = 0$.

(B) The given equation of the circle is $x^2 + y^2 + 6y = 0$.
Completing the square for the $y$-terms:
$x^2 + (y^2 + 6y + 9) = 9$
$x^2 + (y + 3)^2 = 3^2$
Comparing this with the standard form $(x - h)^2 + (y - k)^2 = r^2$,the center is $(0, -3)$ and the radius $r = 3$.
Since the distance from the center $(0, -3)$ to the $x$-axis is $|-3| = 3$,which is equal to the radius,the circle touches the $x$-axis.
Since the $x$-coordinate of the center is $0$,the circle touches the $x$-axis at the origin $(0, 0)$.

(B) The lines $x = a, x = 2a, y = -a, y = a$ form a rectangle with vertices at $(a, -a), (2a, -a), (2a, a),$ and $(a, a)$.
Since the quadrilateral is a rectangle,its circumcircle has the diagonal as its diameter.
Taking the diagonal connecting $(a, -a)$ and $(2a, a)$,the equation of the circle is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the coordinates: $(x - a)(x - 2a) + (y - (-a))(y - a) = 0$.
Expanding this: $(x^2 - 3ax + 2a^2) + (y + a)(y - a) = 0$.
$(x^2 - 3ax + 2a^2) + (y^2 - a^2) = 0$.
$x^2 + y^2 - 3ax + a^2 = 0$.

(A) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the given points $(4, 3)$ and $(-12, -1)$:
$(x - 4)(x - (-12)) + (y - 3)(y - (-1)) = 0$
$(x - 4)(x + 12) + (y - 3)(y + 1) = 0$
Expanding the terms:
$(x^2 + 12x - 4x - 48) + (y^2 + y - 3y - 3) = 0$
$x^2 + y^2 + 8x - 2y - 51 = 0$

(A) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Given endpoints are $(0, 1)$ and $(1, 1)$.
Substituting these values into the formula:
$(x - 0)(x - 1) + (y - 1)(y - 1) = 0$
$x(x - 1) + (y - 1)^2 = 0$
$x^2 - x + y^2 - 2y + 1 = 0$
$x^2 + y^2 - x - 2y + 1 = 0$.

(D) The centre of circle $C$ lies on the line passing through $D$ perpendicular to $PQ$. Thus,the centre of $C$ lies on $y - \frac{3}{2} = \left(\frac{1}{\sqrt{3}}\right)(x - \frac{3\sqrt{3}}{2}) \Rightarrow x = \sqrt{3}y$.
Let the centre of the circle $C$ be $(\sqrt{3}y_1, y_1)$. Then,
$(\frac{3\sqrt{3}}{2} - \sqrt{3}y_1)^2 + (\frac{3}{2} - y_1)^2 = 1$
$\Rightarrow 3(\frac{3}{2} - y_1)^2 + (\frac{3}{2} - y_1)^2 = 1$
$\Rightarrow 4(\frac{3}{2} - y_1)^2 = 1$ $\Rightarrow \frac{3}{2} - y_1 = \pm \frac{1}{2}$ $\Rightarrow y_1 = 1$ or $y_1 = 2$.
Thus,the centre of $C$ can be $(\sqrt{3}, 1)$ or $(2\sqrt{3}, 2)$. Since the centre of the circle and the origin lie on the same side of $\sqrt{3}x + y - 6 = 0$,and $\sqrt{3}(0) + 0 - 6 < 0$ and $\sqrt{3}(\sqrt{3}) + 1 - 6 = 3 + 1 - 6 = -2 < 0$,the centre of the circle $C$ is $(\sqrt{3}, 1)$.
Hence,the equation of the circle is $(x - \sqrt{3})^2 + (y - 1)^2 = 1$.

(A) For the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$,we have $a^2 = 16$ and $b^2 = 9$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$.
The coordinates of the foci are $(\pm ae, 0) = (\pm 4 \times \frac{\sqrt{7}}{4}, 0) = (\pm \sqrt{7}, 0)$.
The circle passes through $(\sqrt{7}, 0)$ and has its center at $(0, 3)$.
The radius $r$ of the circle is the distance between $(0, 3)$ and $(\sqrt{7}, 0)$:
$r^2 = (\sqrt{7} - 0)^2 + (0 - 3)^2 = 7 + 9 = 16$.
The equation of the circle with center $(h, k) = (0, 3)$ and radius $r = 4$ is:
$(x - 0)^2 + (y - 3)^2 = 16$
$x^2 + y^2 - 6y + 9 = 16$
$x^2 + y^2 - 6y - 7 = 0$.

(D) The center of the circle is the point of intersection of the two diameters $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$.
Solving these equations: $3x - 4y = 7$ and $2x - 3y = 5$.
Multiplying the first by $3$ and the second by $4$: $9x - 12y = 21$ and $8x - 12y = 20$.
Subtracting gives $x = 1$. Substituting $x = 1$ into $2x - 3y = 5$ gives $2 - 3y = 5$,so $y = -1$.
The center is $(1, -1)$.
The area of the circle is $\pi r^2 = 49\pi$,so $r^2 = 49$ and $r = 7$.
The equation of the circle is $(x - 1)^2 + (y + 1)^2 = 7^2$.
Expanding this: $x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.
Thus,$x^2 + y^2 - 2x + 2y - 47 = 0$.

(B) The equation of a circle with center $(h, k)$ tangent to the $x$-axis is $(x-h)^{2} + (y-k)^{2} = k^{2}$.
Since the circle passes through the point $(-1, 1)$,we substitute these coordinates into the equation:
$(-1-h)^{2} + (1-k)^{2} = k^{2}$
$1 + 2h + h^{2} + 1 - 2k + k^{2} = k^{2}$
$h^{2} + 2h + 2 - 2k = 0$
For $h$ to be a real coordinate,the discriminant $D$ of this quadratic equation in $h$ must be greater than or equal to $0$:
$D = (2)^{2} - 4(1)(2 - 2k) \ge 0$
$4 - 8 + 8k \ge 0$
$8k - 4 \ge 0$
$8k \ge 4$
$k \ge \frac{1}{2}$

(C) Let the equation of the circle be $(x-3)^2 + (y-0)^2 + \lambda y = 0$.
Since it passes through $(1, -2)$:
$(1-3)^2 + (-2)^2 + \lambda(-2) = 0$
$4 + 4 - 2\lambda = 0$
$8 = 2\lambda \Rightarrow \lambda = 4$.
Thus,the equation of the circle is $(x-3)^2 + y^2 + 4y = 0$.
Checking the options:
For $(5, -2)$: $(5-3)^2 + (-2)^2 + 4(-2) = 2^2 + 4 - 8 = 4 + 4 - 8 = 0$.
Therefore,the circle passes through $(5, -2)$.

(A) Given the ellipse equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$,we have $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.
Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.
The foci are at $(\pm ae, 0) = (\pm \sqrt{7}, 0)$.
The circle passes through $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ and has its centre at $(0, 3)$.
The radius $r$ of the circle is the distance from the centre $(0, 3)$ to a focus $(\sqrt{7}, 0)$:
$r^2 = (\sqrt{7} - 0)^2 + (0 - 3)^2 = 7 + 9 = 16$,so $r = 4$.
The equation of the circle with centre $(0, 3)$ and radius $r = 4$ is $(x - 0)^2 + (y - 3)^2 = 4^2$.
$x^2 + y^2 - 6y + 9 = 16$.
$x^2 + y^2 - 6y - 7 = 0$.

(B) Since the circle lies in the second quadrant and touches both axes,its centre must be $(-r, r)$,where $r$ is the radius.
Given $r = 4$,the centre is $(-4, 4)$.
The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - (-4))^2 + (y - 4)^2 = 4^2$.
$(x + 4)^2 + (y - 4)^2 = 16$.
Expanding this,we get $(x^2 + 8x + 16) + (y^2 - 8y + 16) = 16$.
$x^2 + y^2 + 8x - 8y + 16 = 0$.

(A) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Comparing the given equation $x^2 + y^2 + 2x + 4y + 1 = 0$ with the general form:
$2g = 2 \implies g = 1$
$2f = 4 \implies f = 2$
$c = 1$
The centre of the circle is $(-g, -f) = (-1, -2)$.
The radius of the circle is $\sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 2^2 - 1} = \sqrt{1 + 4 - 1} = \sqrt{4} = 2$.
Therefore,the correct option is $A$.

(C) Let the equation of the circle be $x^2 + y^2 = r^2$. Given the radius $r = 4$,the equation is $x^2 + y^2 = 16$.
Since the points $(t, 1/t)$ lie on the circle,they must satisfy the equation:
$t^2 + (1/t)^2 = 16$
Multiplying by $t^2$,we get $t^4 - 16t^2 + 1 = 0$.
This is a biquadratic equation in $t$ where the roots are $a, b, c, d$.
The product of the roots of the equation $t^4 + 0t^3 - 16t^2 + 0t + 1 = 0$ is given by the constant term divided by the leading coefficient.
Thus,$abcd = 1/1 = 1$.

(C) The reflection of the point $(3, 4)$ in the line $y = x$ is obtained by swapping the coordinates,which gives the center of the circle as $(h, k) = (4, 3)$.
Since the circle is tangent to the $y$-axis,the radius $r$ of the circle is equal to the absolute value of the $x$-coordinate of the center,so $r = |h| = 4$.
The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - 4)^2 + (y - 3)^2 = 4^2$.
Expanding this,we have $x^2 - 8x + 16 + y^2 - 6y + 9 = 16$.
Simplifying,we get $x^2 + y^2 - 8x - 6y + 9 = 0$.

(B) The center of the circle is the midpoint of the diameter: $(\frac{3-5}{2}, \frac{8+2}{2}) = (-1, 5)$.
The radius squared $r^2$ is the distance from the center $(-1, 5)$ to $(3, 8)$ squared: $r^2 = (3 - (-1))^2 + (8 - 5)^2 = 4^2 + 3^2 = 16 + 9 = 25$.
The equation of the circle is $(x + 1)^2 + (y - 5)^2 = 25$.
Substitute the point $(k, 10)$ into the equation: $(k + 1)^2 + (10 - 5)^2 = 25$.
$(k + 1)^2 + 5^2 = 25$.
$(k + 1)^2 + 25 = 25$.
$(k + 1)^2 = 0$.
$k + 1 = 0$,so $k = -1$.
Since $k = -1$ is an integer,there is exactly one integral value of $k$.

(B) In an equilateral triangle,the circumcenter coincides with the centroid. The coordinates of the vertices are given as $A(-1 + a \cos \theta, 2 + a \sin \theta),$ $B(-1 + a \cos \alpha, 2 - a \sin \alpha),$ and $C(-1 + a \sin \beta, 2 + a \cos \beta).$
By observing the structure,the circumcenter is $O(-1, 2).$
The distance from the circumcenter to any vertex is the circumradius $R.$
In an equilateral triangle,the length of the median $m = \frac{3}{2}R.$
Given $m = 2b,$ we have $2b = \frac{3}{2}R \implies R = \frac{4b}{3}.$
The equation of the circle with center $(-1, 2)$ and radius $R = \frac{4b}{3}$ is $(x + 1)^2 + (y - 2)^2 = (\frac{4b}{3})^2.$
Expanding this: $x^2 + 2x + 1 + y^2 - 4y + 4 = \frac{16b^2}{9}.$
Multiplying by $9$: $9x^2 + 9y^2 + 18x - 36y + 45 = 16b^2.$
Wait,re-evaluating the median length: The distance from $A$ to the midpoint of $BC$ is $2b.$ For an equilateral triangle,$R = \frac{2}{3} \times \text{median} = \frac{2}{3}(2b) = \frac{4b}{3}.$
Actually,the standard relation is $R = \frac{2}{3} \times \text{median}.$ Thus $R = \frac{4b}{3}.$
$(x+1)^2 + (y-2)^2 = \frac{16b^2}{9} \implies 9x^2 + 9y^2 + 18x - 36y + 45 - 16b^2 = 0.$
Comparing with options,option $B$ is correct.