Equations of circle Questions in English

(C) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$.
Since its centre $(-g, -f)$ lies on the line $x-4y=1$,we have $-g-4(-f)=1$,which simplifies to $-g+4f=1$ $\dots(i)$.
Since the circle passes through $(3,7)$,we have $3^2+7^2+2g(3)+2f(7)+c=0$,which gives $6g+14f+c=-58$ $\dots(ii)$.
Since the circle passes through $(5,5)$,we have $5^2+5^2+2g(5)+2f(5)+c=0$,which gives $10g+10f+c=-50$ $\dots(iii)$.
Subtracting equation $(ii)$ from $(iii)$,we get $(10g-6g)+(10f-14f) = -50 - (-58)$,so $4g-4f=8$,or $g-f=2$ $\dots(iv)$.
Solving $(i)$ and $(iv)$: adding them gives $3f=3$,so $f=1$. Substituting $f=1$ in $(iv)$,we get $g=3$.
Substituting $g=3$ and $f=1$ in $(iii)$,we get $10(3)+10(1)+c=-50$,so $30+10+c=-50$,which gives $c=-90$.
Thus,the equation of the circle is $x^2+y^2+6x+2y-90=0$.

(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$.
Since it passes through $(1, -2)$,we have $1+4+2g-4f+c=0 \Rightarrow 2g-4f+c=-5$ $(i)$.
Since it passes through $(4, -3)$,we have $16+9+8g-6f+c=0 \Rightarrow 8g-6f+c=-25$ $(ii)$.
Subtracting $(i)$ from $(ii)$: $(8g-6f+c) - (2g-4f+c) = -25 - (-5) \Rightarrow 6g-2f = -20$ $(iii)$.
The centre $(-g, -f)$ lies on $3x+2y=7$,so $3(-g)+2(-f)=7 \Rightarrow -3g-2f=7$ $(iv)$.
Subtracting $(iv)$ from $(iii)$: $(6g-2f) - (-3g-2f) = -20 - 7$ $\Rightarrow 9g = -27$ $\Rightarrow g = -3$.
Substituting $g=-3$ in $(iii)$: $6(-3)-2f = -20$ $\Rightarrow -18-2f = -20$ $\Rightarrow -2f = -2$ $\Rightarrow f = 1$.
Substituting $g=-3$ and $f=1$ in $(i)$: $2(-3)-4(1)+c = -5$ $\Rightarrow -6-4+c = -5$ $\Rightarrow c = 5$.
The equation of the circle is $x^2+y^2-6x+2y+5=0$.

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

(A) The center of the circle is the point of intersection of the diameters $2x - 3y = 5$ and $3x - 4y = 7$.
Solving these equations:
Multiply the first by $3$ and the second by $2$: $6x - 9y = 15$ and $6x - 8y = 14$.
Subtracting the equations gives $y = -1$. Substituting $y = -1$ into $2x - 3(-1) = 5$ gives $2x + 3 = 5$,so $x = 1$.
The center is $(1, -1)$.
The area of the circle is $\pi r^2 = 154$.
Using $\pi = \frac{22}{7}$,we have $\frac{22}{7} r^2 = 154$,which gives $r^2 = 154 \times \frac{7}{22} = 49$,so $r = 7$.
The equation of the circle is $(x - 1)^2 + (y + 1)^2 = 7^2$.
Expanding this: $x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.
$x^2 + y^2 - 2x + 2y - 47 = 0$.

(B) Given the equation of the circle: $x^2+y^2-6x-2y+9=0$.
Completing the square for $x$ and $y$ terms:
$(x^2-6x+9) + (y^2-2y+1) = -9+9+1$.
$(x-3)^2 + (y-1)^2 = 1^2$.
The standard form of a circle is $(x-h)^2 + (y-k)^2 = r^2$,where the center is $(h, k) = (3, 1)$ and the radius $r = 1$.
The parametric equations are given by $x = h + r \cos \theta$ and $y = k + r \sin \theta$.
Substituting the values: $x = 3 + 1 \cos \theta$ and $y = 1 + 1 \sin \theta$.
Thus,$x = 3 + \cos \theta$ and $y = 1 + \sin \theta$.

(D) The given equation is $x^2+y^2-ax-by=0$.
Completing the square,we get:
$(x-\frac{a}{2})^2 + (y-\frac{b}{2})^2 = \frac{a^2+b^2}{4}$.
This represents a circle with center $(h, k) = (\frac{a}{2}, \frac{b}{2})$ and radius $r = \sqrt{\frac{a^2+b^2}{4}}$.
The parametric equations of a circle are $x = h + r \cos \theta$ and $y = k + r \sin \theta$.
Substituting the values,we get $x = \frac{a}{2} + \sqrt{\frac{a^2+b^2}{4}} \cos \theta$ and $y = \frac{b}{2} + \sqrt{\frac{a^2+b^2}{4}} \sin \theta$.

(D) Let the centre of the circle be $(h, k)$. Since it lies on the line $x-4y=1$,we have $h = 1+4k$. Thus,the centre is $(4k+1, k)$.
Since the circle passes through $(3,7)$ and $(5,5)$,the distances from the centre to these points are equal (equal to the radius $r$):
$(4k+1-3)^2 + (k-7)^2 = (4k+1-5)^2 + (k-5)^2$
$(4k-2)^2 + (k-7)^2 = (4k-4)^2 + (k-5)^2$
$16k^2 - 16k + 4 + k^2 - 14k + 49 = 16k^2 - 32k + 16 + k^2 - 10k + 25$
$17k^2 - 30k + 53 = 17k^2 - 42k + 41$
$12k = -12 \Rightarrow k = -1$.
Substituting $k = -1$ into $h = 1+4k$,we get $h = 1+4(-1) = -3$. So,the centre is $(-3, -1)$.
The radius $r$ is the distance between $(-3, -1)$ and $(5, 5)$:
$r^2 = (5 - (-3))^2 + (5 - (-1))^2 = 8^2 + 6^2 = 64 + 36 = 100$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$:
$(x + 3)^2 + (y + 1)^2 = 100$
$x^2 + 6x + 9 + y^2 + 2y + 1 = 100$
$x^2 + y^2 + 6x + 2y - 90 = 0$.

(A) Given the circumference of the circle is $10 \pi$ units.
We know that the circumference $C = 2 \pi r$,so $2 \pi r = 10 \pi$,which gives $r = 5$.
The center of the circle is $(h, k) = (2, -3)$.
The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - 2)^2 + (y - (-3))^2 = 5^2$.
$(x - 2)^2 + (y + 3)^2 = 25$.
Expanding this,we get $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 25$.
$x^2 + y^2 - 4x + 6y + 13 = 25$.
$x^2 + y^2 - 4x + 6y - 12 = 0$.

(C) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$.
Since the circle passes through the origin $(0,0)$,we have $c=0$.
The circle cuts off an intercept of $-2$ on the $X$-axis,so it passes through $(-2,0)$. Substituting this into the equation: $(-2)^2 + 0^2 + 2g(-2) + 2f(0) + 0 = 0$ $\Rightarrow 4 - 4g = 0$ $\Rightarrow g = 1$.
The circle cuts off an intercept of $3$ on the $Y$-axis,so it passes through $(0,3)$. Substituting this into the equation: $0^2 + 3^2 + 2g(0) + 2f(3) + 0 = 0$ $\Rightarrow 9 + 6f = 0$ $\Rightarrow f = -\frac{3}{2}$.
Substituting $g=1$ and $f=-\frac{3}{2}$ into the general equation: $x^2+y^2+2(1)x+2(-\frac{3}{2})y+0=0$.
This simplifies to $x^2+y^2+2x-3y=0$.

(B) The given equation of the circle is $x^{2}+y^{2}-6x+4y-3=0$.
Completing the square,we get:
$(x^{2}-6x+9) + (y^{2}+4y+4) = 3+9+4$
$(x-3)^{2} + (y+2)^{2} = 16$
$(x-3)^{2} + (y+2)^{2} = 4^{2}$
Comparing this with the standard form $(x-h)^{2} + (y-k)^{2} = r^{2}$,we identify the center $(h, k) = (3, -2)$ and radius $r = 4$.
The parametric equations of a circle are given by $x = h + r \cos \theta$ and $y = k + r \sin \theta$.
Substituting the values,we get $x = 3 + 4 \cos \theta$ and $y = -2 + 4 \sin \theta$.

(D) The general equation of a circle is $x^{2}+y^{2}+2gx+2fy+c=0$,with center $(-g, -f)$.
For the first circle $x^{2}+y^{2}+2x-4y+1=0$,the center $A$ is $(-1, 2)$.
For the second circle $x^{2}+y^{2}-8x+6y+17=0$,the center $B$ is $(4, -3)$.
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 coordinates of $A$ and $B$:
$(x-(-1))(x-4)+(y-2)(y-(-3))=0$
$(x+1)(x-4)+(y-2)(y+3)=0$
$x^{2}-4x+x-4+y^{2}+3y-2y-6=0$
$x^{2}+y^{2}-3x+y-10=0$.

(C) Since the circle passes through the origin $(0,0)$ and makes intercepts of $3$ on the $x$-axis and $-5$ on the $y$-axis,the points $(3,0)$ and $(0,-5)$ lie on the circle.
Since the angle subtended by the diameter at any point on the circle is $90^{\circ}$,the line segment joining $(3,0)$ and $(0,-5)$ acts as the diameter of the circle because the angle at the origin $(0,0)$ is $90^{\circ}$.
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 points $(3,0)$ and $(0,-5)$:
$(x-3)(x-0) + (y-0)(y-(-5)) = 0$
$x(x-3) + y(y+5) = 0$
$x^{2} - 3x + y^{2} + 5y = 0$
$x^{2} + y^{2} - 3x + 5y = 0$

(D) The general equation of a circle is $x^{2}+y^{2}+2gx+2fy+c=0$,where the radius $r = \sqrt{g^{2}+f^{2}-c}$.
Comparing the given equation $x^{2}+y^{2}-4x+6y-k=0$ with the general form:
$2g = -4 \Rightarrow g = -2$
$2f = 6 \Rightarrow f = 3$
$c = -k$
Given the radius $r = 5$,we have:
$5 = \sqrt{(-2)^{2} + (3)^{2} - (-k)}$
$5 = \sqrt{4 + 9 + k}$
Squaring both sides:
$25 = 13 + k$
$k = 25 - 13 = 12$

(A) The center of the first circle $x^{2}+y^{2}-2x+3y-3=0$ is found by comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,giving $C_{1} = (-g, -f) = (1, -3/2)$.
The center of the second circle $x^{2}+y^{2}+6x-12y-5=0$ is $C_{2} = (-3, 6)$.
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 centers $(1, -3/2)$ and $(-3, 6)$:
$(x-1)(x+3) + (y+3/2)(y-6) = 0$
$x^{2}+2x-3 + y^{2}-6y+3/2y-9 = 0$
$x^{2}+y^{2}+2x-9/2y-12 = 0$
Multiplying by $2$ to match the options:
$2x^{2}+2y^{2}+4x-9y-24 = 0$.

(B) The vertices of the rectangle are $(a, b)$,$(-a, b)$,$(-a, -b)$,and $(a, -b)$.
Since the circle passes through these vertices,the diagonal of the rectangle acts as the diameter of the circle.
Taking the diagonal connecting $(a, b)$ and $(-a, -b)$ as the diameter,the equation of the circle is given by:
$(x - a)(x + a) + (y - b)(y + b) = 0$
$x^2 - a^2 + y^2 - b^2 = 0$
$x^2 + y^2 = a^2 + b^2$

(B) Let the equation of the circle be $x^{2} + y^{2} + 2gx + 2fy + k = 0$.
Since it passes through $(2,0)$,we have $4 + 4g + k = 0 \Rightarrow k = -4 - 4g$.
Since it passes through $(0,1)$,we have $1 + 2f + k = 0 \Rightarrow k = -1 - 2f$.
Since it passes through $(4,5)$,we have $16 + 25 + 8g + 10f + k = 0 \Rightarrow 41 + 8g + 10f + k = 0$.
Solving these equations,we get $g = -\frac{13}{6}$,$f = -\frac{17}{6}$,and $k = \frac{14}{3}$.
The equation of the circle is $x^{2} + y^{2} - \frac{13}{3}x - \frac{17}{3}y + \frac{14}{3} = 0$,which simplifies to $3(x^{2} + y^{2}) - 13x - 17y + 14 = 0$.
Since $(0, c)$ lies on this circle,we substitute $x=0$ and $y=c$:
$3(0^{2} + c^{2}) - 13(0) - 17(c) + 14 = 0$.
$3c^{2} - 17c + 14 = 0$.
$(3c - 14)(c - 1) = 0$.
Thus,$c = 1$ or $c = \frac{14}{3}$.
Since $(0,1)$ is already a given point,the other point is $(0, \frac{14}{3})$,so $c = \frac{14}{3}$.

(C) The given circle is $2x^2+2y^2-6x+8y+1=0$.
Dividing by $2$,we get $x^2+y^2-3x+4y+\frac{1}{2}=0$.
The center is $(\frac{3}{2}, -2)$ and the radius $r_1$ is given by $r_1^2 = (\frac{3}{2})^2 + (-2)^2 - \frac{1}{2} = \frac{9}{4} + 4 - \frac{1}{2} = \frac{9+16-2}{4} = \frac{23}{4}$.
Let the required circle be $x^2+y^2-3x+4y+k=0$.
Its radius $r_2$ satisfies $r_2^2 = (\frac{3}{2})^2 + (-2)^2 - k = \frac{25}{4} - k$.
Since the area is double,$\pi r_2^2 = 2(\pi r_1^2)$,so $r_2^2 = 2r_1^2$.
$\frac{25}{4} - k = 2(\frac{23}{4}) = \frac{23}{2}$.
$k = \frac{25}{4} - \frac{46}{4} = -\frac{21}{4}$.
Substituting $k$ back,$x^2+y^2-3x+4y-\frac{21}{4}=0$.
Multiplying by $4$,we get $4x^2+4y^2-12x+16y-21=0$.

(A) The given equation of the circle is $x^2+y^2-6x-4y-12=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2$.
The center of the circle is $(-g, -f) = (3, 2)$.
$A$ circle concentric with the given circle will have the same center $(3, 2)$.
Let the equation of the required circle be $(x-3)^2+(y-2)^2=r^2$.
Since the circle touches the $Y$-axis,the radius $r$ is equal to the absolute value of the $x$-coordinate of the center,so $r = |3| = 3$.
Substituting $r=3$ into the equation: $(x-3)^2+(y-2)^2=3^2$.
Expanding this,we get $x^2-6x+9+y^2-4y+4=9$.
Simplifying,we get $x^2+y^2-6x-4y+4=0$.

(A) The radius $r$ of the circle is the perpendicular distance from the center $(3, 4)$ to the line $5x + 12y - 11 = 0$.
$r = \left| \frac{5(3) + 12(4) - 11}{\sqrt{5^2 + 12^2}} \right| = \left| \frac{15 + 48 - 11}{\sqrt{25 + 144}} \right| = \frac{52}{13} = 4$.
The equation of the circle with center $(h, k) = (3, 4)$ and radius $r = 4$ is given by $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 3)^2 + (y - 4)^2 = 4^2$
$x^2 - 6x + 9 + y^2 - 8y + 16 = 16$
$x^2 + y^2 - 6x - 8y + 9 = 0$.

(A) Given equations are $x = 6 \cos \theta$ and $y = 6 \sin \theta$.
Squaring both equations,we get $x^{2} = 36 \cos^{2} \theta$ and $y^{2} = 36 \sin^{2} \theta$.
Adding these two equations:
$x^{2} + y^{2} = 36 \cos^{2} \theta + 36 \sin^{2} \theta$
$x^{2} + y^{2} = 36(\cos^{2} \theta + \sin^{2} \theta)$
Since $\cos^{2} \theta + \sin^{2} \theta = 1$,we have:
$x^{2} + y^{2} = 36$.

(B) The given parametric equations are $x = 4a \left(\frac{1-t^2}{1+t^2}\right)$ and $y = \frac{8at}{1+t^2}$.
Let $t = \tan \theta$. Then $\cos 2\theta = \frac{1-t^2}{1+t^2}$ and $\sin 2\theta = \frac{2t}{1+t^2}$.
Substituting these,we get $x = 4a \cos 2\theta$ and $y = 4a \sin 2\theta$.
Squaring and adding both equations:
$x^2 + y^2 = (4a \cos 2\theta)^2 + (4a \sin 2\theta)^2$
$x^2 + y^2 = 16a^2 (\cos^2 2\theta + \sin^2 2\theta)$
$x^2 + y^2 = (4a)^2$.
This represents a circle with centre $(0,0)$ and radius $4a$.

(D) Given equations are $x=3+5 \cos \theta$ and $y=2+5 \sin \theta$.
Rearranging the terms,we get $\frac{x-3}{5} = \cos \theta$ and $\frac{y-2}{5} = \sin \theta$.
Using the identity $\cos^{2} \theta + \sin^{2} \theta = 1$,we substitute the expressions:
$(\frac{x-3}{5})^{2} + (\frac{y-2}{5})^{2} = 1$.
Expanding the squares: $\frac{x^{2}-6x+9}{25} + \frac{y^{2}-4y+4}{25} = 1$.
Multiplying by $25$: $x^{2}-6x+9 + y^{2}-4y+4 = 25$.
Simplifying: $x^{2}+y^{2}-6x-4y+13 = 25$.
Thus,the Cartesian equation is $x^{2}+y^{2}-6x-4y-12 = 0$.

(C) Given the equation of the circle: $x^2+y^2+ax+by=0$.
Completing the square for $x$ and $y$ terms:
$(x^2+ax+\frac{a^2}{4}) + (y^2+by+\frac{b^2}{4}) = \frac{a^2}{4} + \frac{b^2}{4}$.
This simplifies to: $(x+\frac{a}{2})^2 + (y+\frac{b}{2})^2 = \frac{a^2+b^2}{4}$.
Comparing this with the standard form of a circle $(x-h)^2 + (y-k)^2 = r^2$,we identify the center $(h, k) = (-\frac{a}{2}, -\frac{b}{2})$ and the radius $r = \sqrt{\frac{a^2+b^2}{4}}$.
The parametric equations for a circle with center $(h, k)$ and radius $r$ are given by $x = h + r \cos \theta$ and $y = k + r \sin \theta$.
Substituting the values,we get $x = -\frac{a}{2} + \sqrt{\frac{a^2+b^2}{4}} \cos \theta$ and $y = -\frac{b}{2} + \sqrt{\frac{a^2+b^2}{4}} \sin \theta$.

(A) Given the equation of the line: $2x + 3y - k = 0, k > 0$.
Rewriting in intercept form: $\frac{x}{k/2} + \frac{y}{k/3} = 1$.
Thus,the coordinates of $A$ and $B$ are $(\frac{k}{2}, 0)$ and $(0, \frac{k}{3})$ respectively.
The area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{k}{2} \times \frac{k}{3} = \frac{k^{2}}{12}$.
Given area $= 12$,so $\frac{k^{2}}{12} = 12$ $\Rightarrow k^{2} = 144$ $\Rightarrow k = 12$ (since $k > 0$).
Therefore,$A = (6, 0)$ and $B = (0, 4)$.
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 points $(6, 0)$ and $(0, 4)$:
$(x - 6)(x - 0) + (y - 0)(y - 4) = 0$
$x(x - 6) + y(y - 4) = 0$
$x^{2} - 6x + y^{2} - 4y = 0$
$x^{2} + y^{2} - 6x - 4y = 0$.

(B) The smallest circle passing through two points $A(x_1, y_1)$ and $B(x_2, y_2)$ has the line segment $AB$ as its diameter.
Given points are $A(2,2)$ and $B(3,3)$.
The equation of the circle with diameter $AB$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Substituting the coordinates,we get $(x-2)(x-3) + (y-2)(y-3) = 0$.
Expanding this,we have $(x^2 - 5x + 6) + (y^2 - 5y + 6) = 0$.
Simplifying,we get $x^2 + y^2 - 5x - 5y + 12 = 0$.
Thus,option $B$ is correct.

(A) Let the center of the circle be $(h, k)$. Since the circle touches the $Y$-axis at $(0,3)$,the radius $r = |h|$. Thus,the center is $(\pm r, 3)$.
The equation of the circle is $(x \mp r)^{2} + (y-3)^{2} = r^{2}$.
Since the circle makes an intercept of $8$ units on the $X$-axis,the length of the intercept is $2\sqrt{g^{2}-c}$ or using the geometry,$2\sqrt{r^{2}-k^{2}} = 8$.
Here,$k=3$,so $2\sqrt{r^{2}-3^{2}} = 8 \implies \sqrt{r^{2}-9} = 4 \implies r^{2}-9 = 16 \implies r^{2} = 25 \implies r = 5$.
The center is $(\pm 5, 3)$.
The equation is $(x \mp 5)^{2} + (y-3)^{2} = 5^{2}$.
$x^{2} \mp 10x + 25 + y^{2} - 6y + 9 = 25$.
$x^{2} + y^{2} \mp 10x - 6y + 9 = 0$.

(D) The equation of the circle passing through the points $(1,0), (0,1)$ and $(0,0)$ is given by the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.
Substituting $(0,0)$,we get $c = 0$.
Substituting $(1,0)$,we get $1 + 2g = 0 \Rightarrow g = -\frac{1}{2}$.
Substituting $(0,1)$,we get $1 + 2f = 0 \Rightarrow f = -\frac{1}{2}$.
Thus,the equation of the circle is $x^2 + y^2 - x - y = 0$.
Since the point $(2k, 3k)$ lies on this circle,it must satisfy the equation:
$(2k)^2 + (3k)^2 - (2k) - (3k) = 0$
$4k^2 + 9k^2 - 5k = 0$
$13k^2 - 5k = 0$
$k(13k - 5) = 0$
Given $k \neq 0$,we have $13k - 5 = 0$,which implies $k = \frac{5}{13}$.

(A) Given the equation of the circle $x^{2}+y^{2}=1$,the center is $O(0,0)$ and the radius $r_{1} = 1$.
Let the center of the required circle be $C(4,3)$ and its radius be $r_{2}$.
The distance between the centers $O$ and $C$ is $OC = \sqrt{(4-0)^{2} + (3-0)^{2}} = \sqrt{16+9} = \sqrt{25} = 5$.
Since the circles touch externally,the distance between their centers is equal to the sum of their radii:
$OC = r_{1} + r_{2}$
$5 = 1 + r_{2}$
$r_{2} = 4$.
The equation of the circle with center $(h,k) = (4,3)$ and radius $r = 4$ is:
$(x-4)^{2} + (y-3)^{2} = 4^{2}$
$x^{2} - 8x + 16 + y^{2} - 6y + 9 = 16$
$x^{2} + y^{2} - 8x - 6y + 9 = 0$.

(B) Given that,
$x = 2 + 3 \cos \theta \implies x - 2 = 3 \cos \theta \implies \cos \theta = \frac{x - 2}{3}$
$y = 1 - 3 \sin \theta \implies y - 1 = -3 \sin \theta \implies \sin \theta = \frac{y - 1}{-3}$
We know the identity $\sin^2 \theta + \cos^2 \theta = 1$.
Substituting the values,we get:
$\left(\frac{y - 1}{-3}\right)^2 + \left(\frac{x - 2}{3}\right)^2 = 1$
$\frac{(y - 1)^2}{9} + \frac{(x - 2)^2}{9} = 1$
$(x - 2)^2 + (y - 1)^2 = 9$
Comparing this with the standard equation of a circle $(x - h)^2 + (y - k)^2 = r^2$,we get the centre $(h, k) = (2, 1)$ and radius $r = \sqrt{9} = 3$.

(C) The standard equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^{2} + (y-k)^{2} = r^{2}$.
Given center $(h, k) = (-a, -b)$ and radius $r = \sqrt{a^{2}-b^{2}}$.
Substituting these values into the formula:
$(x - (-a))^{2} + (y - (-b))^{2} = (\sqrt{a^{2}-b^{2}})^{2}$
$(x+a)^{2} + (y+b)^{2} = a^{2}-b^{2}$
Expanding the squares:
$x^{2} + 2ax + a^{2} + y^{2} + 2by + b^{2} = a^{2} - b^{2}$
Subtracting $a^{2}$ from both sides and adding $b^{2}$ to both sides:
$x^{2} + y^{2} + 2ax + 2by + b^{2} + b^{2} = 0$
$x^{2} + y^{2} + 2ax + 2by + 2b^{2} = 0$.

(D) Given equation is $x^2+y^2-6x+10y-2=0$.
Completing the square for $x$ and $y$ terms:
$(x^2-6x+9) + (y^2+10y+25) - 9 - 25 - 2 = 0$.
$(x-3)^2 + (y+5)^2 = 36$.
To transform the origin to $(3, -5)$,we substitute $x = X+3$ and $y = Y-5$,which implies $X = x-3$ and $Y = y+5$.
Substituting these into the equation,we get $X^2 + Y^2 = 36$.
Thus,the new equation is $x^2+y^2=36$.

(C) Let the equilateral triangle be $\triangle ABC$ and its median be $AD = 9$ units.
In an equilateral triangle,the centroid $O$ divides the median $AD$ in the ratio $2:1$.
Since the circle is centered at the origin $O(0,0)$ and passes through the vertices,$O$ is the circumcentre of $\triangle ABC$.
In an equilateral triangle,the circumcentre and the centroid coincide.
Therefore,the distance from the centre $O$ to the vertex $A$ is the radius $R$ of the circle.
$R = AO = \frac{2}{3} AD = \frac{2}{3} \times 9 = 6$ units.
The equation of a circle with centre at the origin $(0,0)$ and radius $R$ is $x^2 + y^2 = R^2$.
Substituting $R = 6$,we get $x^2 + y^2 = 6^2 = 36$.

(A) Since $\angle APB = 90^{\circ}$,the point $P$ lies on a circle with diameter $AB$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Here,$(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (-1, 1)$.
Substituting these values into the formula:
$(x-2)(x+1) + (y-3)(y-1) = 0$
$x^2 + x - 2x - 2 + y^2 - y - 3y + 3 = 0$
$x^2 + y^2 - x - 4y + 1 = 0$

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Substituting the points $(1,1), (-6,0),$ and $(-2,2)$ into the equation:
$1$) For $(1,1): 1 + 1 + 2g + 2f + c = 0 \implies 2g + 2f + c = -2$.
$2$) For $(-6,0): 36 + 0 - 12g + 0 + c = 0 \implies -12g + c = -36$.
$3$) For $(-2,2): 4 + 4 - 4g + 4f + c = 0 \implies -4g + 4f + c = -8$.
Subtracting $(1)$ from $(3)$: $(-4g - 2g) + (4f - 2f) + (c - c) = -8 - (-2) \implies -6g + 2f = -6 \implies -3g + f = -3 \implies f = 3g - 3$.
From $(2)$,$c = 12g - 36$.
Substitute $f$ and $c$ into $(1)$: $2g + 2(3g - 3) + (12g - 36) = -2 \implies 2g + 6g - 6 + 12g - 36 = -2 \implies 20g = 40 \implies g = 2$.
Then $f = 3(2) - 3 = 3$ and $c = 12(2) - 36 = -12$.
The equation of the circle is $x^2 + y^2 + 4x + 6y - 12 = 0$.
Testing option $(C) (-2,-8)$: $(-2)^2 + (-8)^2 + 4(-2) + 6(-8) - 12 = 4 + 64 - 8 - 48 - 12 = 68 - 68 = 0$.
Thus,the point $(-2,-8)$ lies on the circle.

(A) Given equation: $ax^2+2hxy+by^2+2gx+2fy+c=0$ $(i)$
If equation $(i)$ represents a circle,then $a=b$ and $h=0$.
Substituting these in $(i)$,we get: $bx^2+by^2+2gx+2fy+c=0$.
Dividing by $b$ (since $b \neq 0$): $x^2+y^2+2(\frac{g}{b})x+2(\frac{f}{b})y+\frac{c}{b}=0$.
For a point circle,the radius $r$ must be $0$.
The radius of a circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{g^2+f^2-c}$.
Here,$g' = \frac{g}{b}$,$f' = \frac{f}{b}$,and $c' = \frac{c}{b}$.
So,$r^2 = (\frac{g}{b})^2 + (\frac{f}{b})^2 - \frac{c}{b} = 0$.
This implies $\frac{g^2+f^2}{b^2} = \frac{c}{b}$.
Multiplying by $b^2$,we get $g^2+f^2 = bc$.
Since $g^2+f^2 \geq 0$,it follows that $bc \geq 0$.
For a point circle other than the origin,$g^2+f^2 > 0$,therefore $bc > 0$.

(C) The general equation of a second-degree curve is $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$.
For this to represent a circle,the coefficient of $x^2$ must equal the coefficient of $y^2$ (i.e.,$a = b$) and the coefficient of $xy$ must be zero (i.e.,$h = 0$).
Thus,the equation becomes $a(x^2 + y^2) + 2gx + 2fy + c = 0$.
Since the circle passes through the origin $(0, 0)$,substituting $x = 0$ and $y = 0$ into the equation gives $a(0)^2 + a(0)^2 + 2g(0) + 2f(0) + c = 0$,which implies $c = 0$.
Therefore,the conditions are $a = b, h = 0, c = 0$.

(C) Let the radius of the required circle be $r$. Given that the circumference is $10 \pi$.
$2 \pi r = 10 \pi$
$\Rightarrow r = 5$
The centre of the circle is $(h, k) = (2, -3)$.
The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values:
$(x - 2)^2 + (y - (-3))^2 = 5^2$
$(x - 2)^2 + (y + 3)^2 = 25$
$x^2 - 4x + 4 + y^2 + 6y + 9 = 25$
$x^2 + y^2 - 4x + 6y + 13 - 25 = 0$
$x^2 + y^2 - 4x + 6y - 12 = 0$
Thus,option $C$ is correct.

(C) Since the circle touches both coordinate axes,its center is $(h, h)$ and its radius is $|h|$.
Given that the circle also touches the line $3x - 4y - 12 = 0$,the perpendicular distance from the center $(h, h)$ to the line must equal the radius $|h|$.
$\therefore \left|\frac{3h - 4h - 12}{\sqrt{3^2 + (-4)^2}}\right| = |h|$
$\Rightarrow \left|\frac{-h - 12}{5}\right| = |h|$
$\Rightarrow |-h - 12| = 5|h|$
Case $1$: $-h - 12 = 5h$ $\Rightarrow 6h = -12$ $\Rightarrow h = -2$.
Case $2$: $-h - 12 = -5h$ $\Rightarrow 4h = 12$ $\Rightarrow h = 3$.
For $h = 3$,the equation is $(x - 3)^2 + (y - 3)^2 = 3^2$.
$x^2 - 6x + 9 + y^2 - 6y + 9 = 9$
$x^2 + y^2 - 6x - 6y + 9 = 0$.

(D) The given lines $3x + 4y - 18 = 0$ and $3x + 4y + 2 = 0$ are parallel tangents to the circle.
The distance between these parallel lines is $d = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} = \frac{|2 - (-18)|}{\sqrt{3^2 + 4^2}} = \frac{20}{5} = 4$.
The diameter of the circle is $4$,so the radius $r = \frac{4}{2} = 2$.
Let the centre of the circle be $(h, k)$. Since the centre lies on $2x + y + 3 = 0$,we have $2h + k + 3 = 0 \Rightarrow k = -2h - 3$.
The perpendicular distance from $(h, k)$ to the tangent $3x + 4y + 2 = 0$ is equal to the radius $r = 2$:
$\frac{|3h + 4k + 2|}{\sqrt{3^2 + 4^2}} = 2$ $\Rightarrow |3h + 4(-2h - 3) + 2| = 10$ $\Rightarrow |3h - 8h - 12 + 2| = 10$ $\Rightarrow |-5h - 10| = 10$.
This gives $-5h - 10 = 10 \Rightarrow h = -4$ or $-5h - 10 = -10 \Rightarrow h = 0$.
If $h = -4$,then $k = -2(-4) - 3 = 5$. The centre is $(-4, 5)$.
The equation of the circle is $(x + 4)^2 + (y - 5)^2 = 2^2$ $\Rightarrow x^2 + 8x + 16 + y^2 - 10y + 25 = 4$ $\Rightarrow x^2 + y^2 + 8x - 10y + 37 = 0$.

(C) The given equation of the circle is $2x^2 + 2y^2 = 9$.
Dividing by $2$,we get $x^2 + y^2 = \frac{9}{2}$.
This is in the form $x^2 + y^2 = r^2$,where $r^2 = \frac{9}{2}$,so $r = \frac{3}{\sqrt{2}}$.
The parametric equations for a circle $x^2 + y^2 = r^2$ are given by $x = r \cos \theta$ and $y = r \sin \theta$.
Substituting $r = \frac{3}{\sqrt{2}}$,we get $x = \frac{3}{\sqrt{2}} \cos \theta$ and $y = \frac{3}{\sqrt{2}} \sin \theta$.

(D) The given equation of the circle is $x^2+y^2-2x+ky+4=0$ $(i)$.
The center of the circle $(i)$ is $C \equiv (1, -k/2)$.
Since the circle $S$ is concentric with circle $(i)$,the equation of $S$ is $(x-1)^2 + (y+k/2)^2 = r^2$ $(ii)$.
The center $(1, -k/2)$ lies on the diameter line $3x-2y+4=0$.
Substituting the center coordinates into the line equation: $3(1) - 2(-k/2) + 4 = 0$ $\Rightarrow 3 + k + 4 = 0$ $\Rightarrow k = -7$.
Substituting $k = -7$ into equation $(ii)$,we get $(x-1)^2 + (y-7/2)^2 = r^2$ $(iii)$.
Since the circle $S$ passes through the point $(3, -2)$,we substitute these coordinates into equation $(iii)$:
$(3-1)^2 + (-2-7/2)^2 = r^2$
$2^2 + (-11/2)^2 = r^2$
$4 + 121/4 = r^2$
$r^2 = (16+121)/4 = 137/4$.
Therefore,the radius $r = \frac{\sqrt{137}}{2} = \frac{1}{2}\sqrt{137}$.

(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... $(i)$
The length of the intercept made by the circle on the $X$-axis is $2\sqrt{g^2-c} = \frac{2\sqrt{13}}{3} \Rightarrow g^2-c = \frac{13}{9}$ ... $(ii)$
The length of the intercept made by the circle on the $Y$-axis is $2\sqrt{f^2-c} = \frac{2\sqrt{22}}{3} \Rightarrow f^2-c = \frac{22}{9}$ ... $(iii)$
Given the radius $r = \frac{\sqrt{38}}{3}$,we have $r^2 = g^2+f^2-c = \frac{38}{9}$ ... $(iv)$
From $(ii)$,$g^2 = c + \frac{13}{9}$. From $(iii)$,$f^2 = c + \frac{22}{9}$.
Substituting these into $(iv)$: $(c + \frac{13}{9}) + (c + \frac{22}{9}) - c = \frac{38}{9}$
$c + \frac{35}{9} = \frac{38}{9} \Rightarrow c = \frac{3}{9} = \frac{1}{3}$
Now,$g^2 = \frac{1}{3} + \frac{13}{9} = \frac{16}{9} \Rightarrow g = \pm \frac{4}{3}$
And $f^2 = \frac{1}{3} + \frac{22}{9} = \frac{25}{9} \Rightarrow f = \pm \frac{5}{3}$
Since the centre $C(-g, -f)$ lies in the second quadrant,$g$ must be positive and $f$ must be negative.
Thus,$C = (-\frac{4}{3}, \frac{5}{3})$.

(D) Let the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(2,3)$,we have $(2-h)^2 + (3-k)^2 = r^2$.
The intercept on the line $x=2$ is $2\sqrt{r^2 - (2-h)^2} = 3$. Substituting $r^2 - (2-h)^2 = (3-k)^2$,we get $2\sqrt{(3-k)^2} = 3$,so $|3-k| = 1.5$,which means $k = 4.5$ or $k = 1.5$.
The intercept on the line $y=3$ is $2\sqrt{r^2 - (3-k)^2} = 4$. Substituting $r^2 - (3-k)^2 = (2-h)^2$,we get $2\sqrt{(2-h)^2} = 4$,so $|2-h| = 2$,which means $h = 4$ or $h = 0$.
Since the centre $(h,k)$ is in the first quadrant,$h, k > 0$.
Testing the case $(h,k) = (4, 4.5)$: $r^2 = (2-4)^2 + (3-4.5)^2 = 4 + 2.25 = 6.25$.
The equation is $(x-4)^2 + (y-4.5)^2 = 6.25$,which simplifies to $x^2 - 8x + 16 + y^2 - 9y + 20.25 = 6.25$,or $x^2 + y^2 - 8x - 9y + 30 = 0$.

(A) Let $x_1$ and $x_2$ be the roots of the equation $x^2+2x-a^2=0$. Thus,$(x-x_1)(x-x_2) = x^2+2x-a^2 = 0$.
Similarly,let $y_1$ and $y_2$ be the roots of the equation $y^2+4y-b^2=0$. Thus,$(y-y_1)(y-y_2) = y^2+4y-b^2 = 0$.
Let the coordinates of points $A$ and $B$ be $(x_1, y_1)$ and $(x_2, y_2)$ respectively.
The equation of a circle with diameter $AB$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Substituting the given quadratic expressions:
$(x^2+2x-a^2) + (y^2+4y-b^2) = 0$.
Rearranging the terms,we get:
$x^2+y^2+2x+4y-a^2-b^2=0$.

(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$.
Since the circle touches the $y$-axis at $(0, 4)$,the center is $(\pm r, 4)$ and the radius $r = 4$.
Thus,$g^2 = r^2 = 16$ and $f = \pm 4$.
Since the circle touches the $y$-axis at $(0, 4)$,the constant term $c$ must satisfy $c = f^2 = 16$.
The $x$-intercept is given by $2\sqrt{g^2-c} = 6$,which implies $\sqrt{g^2-c} = 3$,so $g^2-c = 9$.
Substituting $c = 16$,we get $g^2 = 16+9 = 25$,so $g = \pm 5$.
Substituting $g = \pm 5$,$f = \pm 4$,and $c = 16$ into the general equation:
$x^2+y^2 \pm 10x \pm 8y + 16 = 0$.
Given the options,the correct equation is $x^2+y^2 \pm 10x - 8y + 16 = 0$.

(C) The centre of the circle is $(h, k) = (2, -3)$.
Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the centre.
$r = |k| = |-3| = 3$.
The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - 2)^2 + (y - (-3))^2 = 3^2$.
$(x - 2)^2 + (y + 3)^2 = 9$.
Expanding the terms: $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 9$.
$x^2 + y^2 - 4x + 6y + 4 + 9 = 9$.
$x^2 + y^2 - 4x + 6y + 4 = 0$.

(B) The vertices of the square are $A(0, 0)$,$B(a, 0)$,$C(a, a)$,and $D(0, a)$.
Since the square is inscribed in a circle,the diagonal $AC$ (or $BD$) is the diameter of the circle.
Using the diameter form of the circle equation with endpoints $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (a, a)$:
$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$
$(x - 0)(x - a) + (y - 0)(y - a) = 0$
$x(x - a) + y(y - a) = 0$
$x^2 - ax + y^2 - ay = 0$
$x^2 + y^2 - a(x + y) = 0$

(D) Let $C(h, k)$ be the centre of the given circle.
Since $C(h, k)$ lies on the line $x-y-1=0$,we have $h-k-1=0$,which implies $h=k+1$.
Given that the radius is $3$,the distance from $C(h, k)$ to $P(7, 3)$ is $3$.
So,$(h-7)^2 + (k-3)^2 = 3^2 = 9$.
Substituting $h=k+1$,we get $(k+1-7)^2 + (k-3)^2 = 9$.
$(k-6)^2 + (k-3)^2 = 9$.
$k^2 - 12k + 36 + k^2 - 6k + 9 = 9$.
$2k^2 - 18k + 36 = 0$.
$k^2 - 9k + 18 = 0$.
$(k-6)(k-3) = 0$,so $k=6$ or $k=3$.
If $k=6$,then $h=7$,so $C=(7, 6)$. The equation is $(x-7)^2 + (y-6)^2 = 9$,which simplifies to $x^2 - 14x + 49 + y^2 - 12y + 36 = 9$,or $x^2 + y^2 - 14x - 12y + 76 = 0$.
If $k=3$,then $h=4$,so $C=(4, 3)$. The equation is $(x-4)^2 + (y-3)^2 = 9$,which simplifies to $x^2 - 8x + 16 + y^2 - 6y + 9 = 9$,or $x^2 + y^2 - 8x - 6y + 16 = 0$.
Comparing with the given options,$x^2 + y^2 - 14x - 12y + 76 = 0$ is present as option $(d)$.

(C) Let $L_1 \equiv x+y=6$ and $L_2 \equiv x+2y=4$.
The point of intersection of the two diameters $L_1$ and $L_2$ is the center $C$ of the circle.
Solving the system of equations:
$x+y=6 \Rightarrow x=6-y$
Substituting into $L_2$: $(6-y)+2y=4 \Rightarrow y=-2$.
Then $x=6-(-2)=8$.
Thus,the center $C$ is $(8, -2)$.
The circle passes through the point $P(6, 2)$. The radius $r$ is the distance $CP$.
$r^2 = CP^2 = (8-6)^2 + (-2-2)^2 = 2^2 + (-4)^2 = 4 + 16 = 20$.
The equation of the circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
$(x-8)^2 + (y+2)^2 = 20$
$x^2 - 16x + 64 + y^2 + 4y + 4 = 20$
$x^2 + y^2 - 16x + 4y + 48 = 0$.

(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. \\
Since the circle touches the $x$-axis at $(3, 0)$,the center is $(h, k)$ and the radius $r = |k|$. \\
Thus,the equation becomes $(x-h)^2 + (y-k)^2 = k^2$. \\
Since it passes through $(3, 0)$,we have $(3-h)^2 + (0-k)^2 = k^2$ $\Rightarrow (3-h)^2 = 0$ $\Rightarrow h = 3$. \\
Since it also passes through $(1, -2)$,we substitute $h=3$ into the equation: $(1-3)^2 + (-2-k)^2 = k^2$. \\
$(-2)^2 + 4 + 4k + k^2 = k^2$ \\
$4 + 4 + 4k = 0$ $\Rightarrow 4k = -8$ $\Rightarrow k = -2$. \\
Substituting $h=3$ and $k=-2$ into $(x-h)^2 + (y-k)^2 = k^2$: \\
$(x-3)^2 + (y+2)^2 = (-2)^2$ \\
$x^2 - 6x + 9 + y^2 + 4y + 4 = 4$ \\
$x^2 + y^2 - 6x + 4y + 9 = 0$.