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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of the tangent at the point $(0,2)$ to the circle $x^2+y^2-4x+2y-8 = 0$

$\begin{array}{1 1} (A) y = \large\frac{2}{3} x- 2 \\ (B) y = \large\frac{3}{2} x+ 2 \\ (C) y = \large\frac{2}{3} x+ 2 \\ (D) y = \large\frac{1}{3} x-2 \end{array} $

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The general equation of a circle is given by $x^2+y^2+2gx+2fy+c$ where the center is given by $(-g,-f)$ and the radius by $r = \sqrt (g^2+f^2-c)$
Lets start by finding the center of the circle. From the equation, we can determine that $g=-2$ and $f=1 \rightarrow $ center is $(2,-1)$
We can now draw the circle and its tangent at the point $(0,2)$ as follows:
Now, lets take the two points $(x1,y1) = (0,2)$ and $(x2,y2) = (2,-1)$
The gradient of the radius joining these two points is $ m1 = \large\frac{y2-y1}{x2-x1}$$ = \large\frac{-1-2}{2-0}$$ = \large\frac{-3}{2}$
Since the tangent is perpendicular to the radius, we can calculate it's gradient $m2$ as follows:
$m1m2 = -1 \rightarrow m2 = -1 $$\times \large\frac{-2}{3}$$ = \large\frac{2}{3}$
Now, we can find the equation of the tangent: $y - y2 = m2 (x - x2)$
$\qquad y - 2 = \large\frac{2}{3} $$ (x - 0)$
$\qquad y = \large\frac{2}{3} $$x+ 2$
answered Mar 21, 2014 by balaji.thirumalai

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