$\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} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

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$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...