Name: Two Circles Touching Internally and Externally
Uploaded: 2019-07-19
Duration: 6 min 6 s
Description: In this video, we will learn the condition for two circles touching internally and externally using an interesting example. Come, let's learn and watch this...

Question

Find the equation of tangent and normal at

(1, 1)

of the circle

2 x^{2} + 2 y^{2} - 2 x - 5 y + 3 = 0

.

Accepted Answer

Given point is $P (1, 1)$ and circle is $S \equiv 2 x^{2} + 2 y^{2} - 2 x - 5 y + 3 = 0$ , which can be written as

$S \equiv x^{2} + y^{2} - x - \frac{5}{2} y + \frac{3}{2} = 0 \to (i)$

We know that the equation of the tangent at the point $P (x_{1}, y_{1})$ to the circle $S \equiv x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ is $S_{1} \equiv x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$ .

Thus, the equation of tangent at $P (1, 1)$ to the given circle is

$S_{1} = x (1) + y (1) + g (x + 1) + f (y + 1) + c = 0$

Comparing equation $(i)$ with the equation of the circle in general form, we have

$g = \frac{- 1}{2}, f = - \frac{5}{4},$ and $c = \frac{3}{2}$

$∴ S_{1} \equiv x (1) + y (1) + (- \frac{1}{2}) (x + 1) + (- \frac{5}{4}) (y + 1) + \frac{3}{2} = 0$

$\Rightarrow x + y - \frac{1}{2} x - \frac{1}{2} - \frac{5}{4} y - \frac{5}{4} + \frac{3}{2} = 0$

$\Rightarrow \frac{x}{2} - \frac{y}{4} - \frac{1}{4} = 0$

$\Rightarrow 2 x - y - 1 = 0$ , which is the required equation of tangent at $(1, 1)$ to the circle $2 x^{2} + 2 y^{2} - 2 x - 5 y + 3 = 0$ .

Also, we know that the equation of normal at $P (x_{1}, y_{1})$ of the circle $S \equiv x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ is $(x - x_{1}) (y_{1} + f) - (y - y_{1}) (x_{1} + g) = 0$

Therefore, equation of normal at $P (1, 1)$ of the circle $S \equiv x^{2} + y^{2} - x - \frac{5}{2} y + \frac{3}{2} = 0$ is

$(x - 1) (1 - \frac{5}{4}) - (y - 1) (1 - \frac{1}{2}) = 0$

$\Rightarrow (x - 1) (- \frac{1}{4}) - (y - 1) (\frac{1}{2}) = 0$

$\Rightarrow - \frac{x}{4} + \frac{1}{4} - \frac{y}{2} + \frac{1}{2} = 0$

$\Rightarrow - \frac{x}{4} - \frac{y}{2} + \frac{3}{4} = 0$

$\Rightarrow x + 2 y - 3 = 0$

Find the equation of tangent and normal at $(1, 1)$ of the circle $2 x^{2} + 2 y^{2} - 2 x - 5 y + 3 = 0$ .

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the equation of tangent and normal at (1,1) of the circle 2x2+2y2-2x-5y+3=0.

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the equation of tangent and normal at $(1, 1)$ of the circle $2 x^{2} + 2 y^{2} - 2 x - 5 y + 3 = 0$ .