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 pair tangents drawn from

(3, 2)

to the circle

x^{2} + y^{2} - 6 x + 4 y - 2 = 0

.

Accepted Answer

The given equation of the circle is $S = x^{2} + y^{2} - 6 x + 4 y - 2 = 0$ and the point from which a pair of tangents is drawn is $P (3, 2)$ .

Here, the centre of the given circle is at $(3, - 2) and radius r = \sqrt{9 + 4 + 2} = \sqrt{15}$

The equation of the pair of tangents from a point $(x_{1}, y_{1})$ to the circle $S = 0$ is given by $S_{11} S = S_{1}^{2}$

$\Rightarrow (3^{2} + 2^{2} - 6 \cdot 3 + 4 \cdot 2 - 2) (x^{2} + y^{2} - 6 x + 4 y - 2) = {\{x \cdot 3 + y \cdot 2 - 3 (x + 3) + 2 (y + 2) - 2\}}^{2}$

$\Rightarrow x^{2} + y^{2} - 6 x + 4 y - 2 = {(4 y - 7)}^{2}$

$\Rightarrow x^{2} + y^{2} - 6 x + 4 y - 2 = 16 y^{2} + 49 - 56 y$

$\Rightarrow x^{2} - 15 y^{2} - 6 x + 60 y - 51 = 0$

Thus, the equation of the pair of tangents from $(3, 2)$ to the circle $x^{2} + y^{2} - 6 x + 4 y - 2 = 0$ is $x^{2} - 15 y^{2} - 6 x + 60 y - 51 = 0$

Find the pair tangents drawn from $(3, 2)$ to the circle $x^{2} + y^{2} - 6 x + 4 y - 2 = 0$ .

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the pair tangents drawn from 3,2 to the circle x2+y2-6x+4y-2=0.

Important Questions on Circle

Learn and Prepare for any exam you want !

Find the pair tangents drawn from $(3, 2)$ to the circle $x^{2} + y^{2} - 6 x + 4 y - 2 = 0$ .