Tangent and normal to a circle Questions in English

(A) The given curve is $x^2 + y - 7 = 4x$.
To find the equation of the tangent at $(x_1, y_1) = (1, 10)$,we use the transformation rules: $x^2 \to x x_1$ and $y \to \frac{y + y_1}{2}$ and $x \to \frac{x + x_1}{2}$.
Substituting these into the equation:
$x(1) + \frac{y + 10}{2} - 7 = 4 \left( \frac{x + 1}{2} \right)$
$x + \frac{y + 10}{2} - 7 = 2(x + 1)$
Multiply the entire equation by $2$:
$2x + y + 10 - 14 = 4x + 4$
$2x + y - 4 = 4x + 4$
$y = 4x - 2x + 4 + 4$
$y = 2x + 8$

(A) The given equation of the curve is $2y^2 = x + 1$.
Differentiating both sides with respect to $x$,we get:
$4y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{4y}$.
The slope of the tangent at any point $(x, y)$ is $m_t = \frac{1}{4y}$.
The slope of the normal at any point $(x, y)$ is $m_n = -\frac{1}{m_t} = -4y$.
Now,we calculate the slope of the normal at each given point:
$A. (7, 2): m_n = -4(2) = -8$ (Matches $2$).
$B. (0, 1/\sqrt{2}): m_n = -4(1/\sqrt{2}) = -2\sqrt{2}$ (Matches $5$).
$C. (1, -1): m_n = -4(-1) = 4$ (Matches $3$).
$D. (3, \sqrt{2}): m_n = -4(\sqrt{2}) = -4\sqrt{2}$ (Matches $1$).
Thus,the correct matching is $A-2, B-5, C-3, D-1$.

(D) Given curve is $x^2+y^2=16a^2$.
On differentiating the equation with respect to $x$,we get $2x + 2yy' = 0$,which implies $y' = -\frac{x}{y}$.
At the point $(2\sqrt{2}a, 2\sqrt{2}a)$,the slope of the tangent is $m_1 = -\frac{2\sqrt{2}a}{2\sqrt{2}a} = -1$.
The slope of the normal is $m_2 = -\frac{1}{m_1} = 1$.
The equation of the tangent is $y - 2\sqrt{2}a = -1(x - 2\sqrt{2}a)$,which simplifies to $x + y = 4\sqrt{2}a$.
The $X$-intercept of the tangent is found by setting $y=0$,giving $x = 4\sqrt{2}a$. Let this point be $B(4\sqrt{2}a, 0)$.
The equation of the normal is $y - 2\sqrt{2}a = 1(x - 2\sqrt{2}a)$,which simplifies to $y = x$.
The normal passes through the origin $O(0,0)$ and the point $A(2\sqrt{2}a, 2\sqrt{2}a)$.
The triangle is formed by the points $O(0,0)$,$A(2\sqrt{2}a, 2\sqrt{2}a)$,and $B(4\sqrt{2}a, 0)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
Area $= \frac{1}{2} |0(2\sqrt{2}a-0) + 2\sqrt{2}a(0-0) + 4\sqrt{2}a(0-2\sqrt{2}a)| = \frac{1}{2} |4\sqrt{2}a(-2\sqrt{2}a)| = \frac{1}{2} |-16a^2| = 8a^2$.

(B) Given the parametric equations of the curve: $x=2 \sin t$ and $y=2 \cos t$.
Squaring and adding,we get $x^2+y^2=4 \sin^2 t + 4 \cos^2 t = 4$.
This represents a circle with radius $2$.
Differentiating $x^2+y^2=4$ with respect to $x$: $2x + 2y \frac{dy}{dx} = 0$,which gives $\frac{dy}{dx} = -\frac{x}{y}$.
The slope of the tangent at any point $(x, y)$ is $m_T = -\frac{x}{y}$.
The slope of the normal $m_N$ is $-\frac{1}{m_T} = \frac{y}{x}$.
At $t = \frac{\pi}{2}$,the point is $x = 2 \sin(\frac{\pi}{2}) = 2$ and $y = 2 \cos(\frac{\pi}{2}) = 0$.
The slope of the normal at this point is $m_N = \frac{0}{2} = 0$.
The equation of the normal is $y - y_1 = m_N(x - x_1)$.
Substituting the values: $y - 0 = 0(x - 2)$,which simplifies to $y = 0$.

(C) Given the parametric equations of the curve are $x = a(1 + \cos \theta)$ and $y = a \sin \theta$.
First,we find the derivative $\frac{dy}{dx}$:
$\frac{dy}{d\theta} = a \cos \theta$ and $\frac{dx}{d\theta} = -a \sin \theta$.
Thus,$\frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta$.
The slope of the normal at point $\theta$ is $m_n = -\frac{1}{dy/dx} = \frac{1}{\cot \theta} = \tan \theta$.
The equation of the normal at point $(x_1, y_1) = (a(1 + \cos \theta), a \sin \theta)$ is given by:
$y - a \sin \theta = \tan \theta (x - a(1 + \cos \theta))$.
$y - a \sin \theta = \frac{\sin \theta}{\cos \theta} (x - a - a \cos \theta)$.
$y \cos \theta - a \sin \theta \cos \theta = x \sin \theta - a \sin \theta - a \sin \theta \cos \theta$.
$y \cos \theta = x \sin \theta - a \sin \theta$.
$y \cos \theta = (x - a) \sin \theta$.
If we check the point $(a, 0)$,we substitute $x = a$ and $y = 0$ into the equation:
$0 \cdot \cos \theta = (a - a) \sin \theta \Rightarrow 0 = 0$.
Since this holds true for all $\theta$,the normal always passes through the fixed point $(a, 0)$.

(A) Given curves are $x^2+y^2=4$ and $y^2=3x$. Substituting $y^2=3x$ into the first equation:
$x^2+3x-4=0$
$(x+4)(x-1)=0$
Since $x$ must be non-negative for $y^2=3x$,we have $x=1$.
For $x=1$,$y^2=3$,so $y=\sqrt{3}$ (considering the point of intersection in the first quadrant).
Now,differentiate both curves with respect to $x$:
For $x^2+y^2=4$,$2x+2y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$. At $(1, \sqrt{3})$,$m_1 = -\frac{1}{\sqrt{3}}$.
For $y^2=3x$,$2y\frac{dy}{dx}=3 \Rightarrow \frac{dy}{dx} = \frac{3}{2y}$. At $(1, \sqrt{3})$,$m_2 = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
The angle $\theta$ between the curves is given by $\tan \theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$.
$\tan \theta = \left|\frac{-\frac{1}{\sqrt{3}} - \frac{\sqrt{3}}{2}}{1 + (-\frac{1}{\sqrt{3}})(\frac{\sqrt{3}}{2})}\right| = \left|\frac{-\frac{2+3}{2\sqrt{3}}}{1-\frac{1}{2}}\right| = \left|\frac{-\frac{5}{2\sqrt{3}}}{\frac{1}{2}}\right| = \frac{5}{\sqrt{3}}$.

(A) The circle is $x^{2}+y^{2}=4$,so its radius $OA = 2$ and the center $O$ is $(0,0)$.
Since $MA$ is a tangent,$\angle OAM = 90^{\circ}$.
In the right-angled triangle $\triangle OAM$,the hypotenuse $OM = 4$ and the side $OA = 2$.
Using the Pythagorean theorem,$MA = \sqrt{OM^{2} - OA^{2}} = \sqrt{4^{2} - 2^{2}} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$.
The quadrilateral $MAOB$ consists of two congruent right-angled triangles,$\triangle OAM$ and $\triangle OBM$.
Therefore,the area of quadrilateral $MAOB = 2 \times \text{Area}(\triangle OAM)$.
Area $(\triangle OAM) = \frac{1}{2} \times OA \times MA = \frac{1}{2} \times 2 \times 2\sqrt{3} = 2\sqrt{3}$.
Thus,the area of quadrilateral $MAOB = 2 \times 2\sqrt{3} = 4\sqrt{3}$ sq. units.

(A) Given,$y=[|\sin x|+|\cos x|]$ and $x^{2}+y^{2}=10$.
We know that $(|\sin x|+|\cos x|) \in [1, \sqrt{2}]$.
Since $[x]$ is the greatest integer function,$y = [|\sin x|+|\cos x|] = 1$.
The point of intersection of the given curves is found by substituting $y=1$ into $x^{2}+y^{2}=10$:
$x^{2}+1^{2}=10$ $\Rightarrow x^{2}=9$ $\Rightarrow x=\pm 3$.
The points of intersection are $(3, 1)$ and $(-3, 1)$.
For the circle $x^{2}+y^{2}=10$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$.
At point $(-3, 1)$,the slope $m_{1} = -(-3)/1 = 3$.
The curve $y=1$ is a horizontal line,so its slope $m_{2} = 0$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}| = |\frac{3-0}{1+3(0)}| = 3$.
Therefore,$\theta = \tan^{-1}(3)$.

(A) The equation of the circle is $x^{2} + y^{2} = 9$,which has center $(0, 0)$ and radius $r = 3$.
Any line parallel to $3x + 4y = 0$ is of the form $3x + 4y = k$.
The perpendicular distance from the center $(0, 0)$ to the line $3x + 4y - k = 0$ must be equal to the radius $r = 3$.
Using the formula for distance from a point to a line: $\frac{|3(0) + 4(0) - k|}{\sqrt{3^{2} + 4^{2}}} = 3$.
$\frac{|-k|}{\sqrt{9 + 16}} = 3$ $\Rightarrow \frac{|k|}{5} = 3$ $\Rightarrow |k| = 15$.
So,$k = 15$ or $k = -15$.
The lines are $3x + 4y = 15$ and $3x + 4y = -15$.
For the line to touch the circle in the first quadrant,the intercept form $\frac{x}{5} + \frac{y}{3.75} = 1$ (for $k=15$) shows positive intercepts,which lies in the first quadrant.
Thus,the required equation is $3x + 4y = 15$.

(C) Let the equation of the other tangent be $y = mx$,which can be written as $mx - y = 0$.
The center of the circle is $C(2, -1)$. The radius $r$ is the perpendicular distance from $C$ to the tangent $3x + y = 0$.
$r = \frac{|3(2) + 1(-1)|}{\sqrt{3^2 + 1^2}} = \frac{|6 - 1|}{\sqrt{10}} = \frac{5}{\sqrt{10}}$.
Since the other tangent $mx - y = 0$ also touches the circle,the perpendicular distance from $C(2, -1)$ to this line must also be equal to $r$.
$\frac{|m(2) - 1(-1)|}{\sqrt{m^2 + (-1)^2}} = \frac{5}{\sqrt{10}}$
$\frac{|2m + 1|}{\sqrt{m^2 + 1}} = \frac{5}{\sqrt{10}}$
Squaring both sides:
$\frac{(2m + 1)^2}{m^2 + 1} = \frac{25}{10} = \frac{5}{2}$
$2(4m^2 + 4m + 1) = 5(m^2 + 1)$
$8m^2 + 8m + 2 = 5m^2 + 5$
$3m^2 + 8m - 3 = 0$
$(3m - 1)(m + 3) = 0$
So,$m = \frac{1}{3}$ or $m = -3$.
The given tangent is $3x + y = 0$,which has slope $m = -3$.
Therefore,the other tangent has slope $m = \frac{1}{3}$.
The equation is $y = \frac{1}{3}x$,which simplifies to $x - 3y = 0$.

(C) The given equation of the circle is $x^{2}+y^{2}-2x-4y-20=0$.
Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-1, f=-2, c=-20$.
The center $A$ is $(-g, -f) = (1, 2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{1+4+20} = \sqrt{25} = 5$.
The tangent at $B(1, 7)$ is $x(1) + y(7) - (x+1) - 2(y+7) - 20 = 0$,which simplifies to $y=7$.
The tangent at $D(4, -2)$ is $x(4) + y(-2) - (x+4) - 2(y-2) - 20 = 0$,which simplifies to $3x-4y-20=0$.
Solving $y=7$ and $3x-4y-20=0$,we get $3x-28-20=0 \implies 3x=48 \implies x=16$. Thus,$C$ is $(16, 7)$.
The quadrilateral $ABCD$ consists of two congruent triangles $\Delta ABC$ and $\Delta ADC$.
The length of the tangent $BC = \sqrt{(16-1)^{2} + (7-7)^{2}} = 15$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AB = \frac{1}{2} \times 15 \times 5 = 37.5$.
The area of quadrilateral $ABCD = 2 \times \text{Area}(\Delta ABC) = 2 \times 37.5 = 75 \text{ sq units}$.

(A) The given equation of the circle is $x^2+y^2-2x+4y-5=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$ and $f=2$.
The center of the circle $C$ is $(-g, -f) = (1, -2)$.
The normal to a circle at any point on its circumference always passes through the center of the circle.
Therefore,the normal is the line passing through the center $C(1, -2)$ and the given point $A(2, 1)$.
The slope $m$ of the line passing through $(1, -2)$ and $(2, 1)$ is given by $m = \frac{1 - (-2)}{2 - 1} = \frac{3}{1} = 3$.
The equation of the line passing through $(2, 1)$ with slope $m=3$ is $(y - 1) = 3(x - 2)$.
Simplifying this,we get $y - 1 = 3x - 6$,which results in $y = 3x - 5$.