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 all common tangents of the following pairs of circles

x^{2} + y^{2} = 9 and x^{2} + y^{2} - 16 x + 2 y + 49 = 0

Accepted Answer

The given equation of the circles are $S = x^{2} + y^{2} = 9 . . . (i) and S' = x^{2} + y^{2} - 16 x + 2 y + 49 = 0 . . . (ii)$

Centres of the circles $S and S' are C (0, 0) and C' (8, - 1)$ respectively.

The radii of the circles $S and S' are r = \sqrt{9} = 3 and r' = \sqrt{64 + 1 - 49} = 4$

Now, $C C' = \sqrt{{(8 - 0)}^{2} + {(- 1 - 0)}^{2}} = \sqrt{(64 + 1)} = \sqrt{65}$ and $r' + r = 3 + 4 = 7$

Here, $C C' > r + r'$ so, the circles lies outside each other and the external centre of similitude $P (x_{1}, y_{1}) divides C C'$ in the ratio $r : r' = 3 : 4$

$∴ P = (\frac{3 (8) - 4 (0)}{3 - 4}, \frac{3 (- 1) - 4 (0)}{3 - 4}) = (- 24, 3)$

Let $m$ be the slope of direct common tangent and the equation of tangent is $(y - 3) = m (x + 24)$

$\Rightarrow m x - y + 24 m + 3 = 0 . . . (iii)$

As we know that the shortest distance from the centre of circle to the tangent is its radius.

$\Rightarrow r = perpendicular distance from C (0, 0) to (iii)$

$\Rightarrow r = \frac{|m (0) + (- 1) (0) + 24 m + 3|}{\sqrt{m^{2} + {(- 1)}^{2}}}$

$\Rightarrow 3 = \frac{|24 m + 3|}{\sqrt{m^{2} + 1}} \Rightarrow 9 m^{2} + 9 = 576 m^{2} + 144 m + 9$

$\Rightarrow 63 m^{2} + 16 m = 0$

$\Rightarrow m (63 m + 16) = 0$

$∴ m = 0 and \frac{- 16}{63}$

When $m = 0$ , the equation of the required tangent is $x \cdot 0 - y + 24 (0) + 3 = 0 \Rightarrow y - 3 = 0$

When $m = \frac{- 16}{63}$ , the equation of the required tangent is $\frac{- 16 x}{63} - y + (\frac{- 384}{63}) + 3 = 0 \Rightarrow \frac{- 16 x}{63} - y - \frac{195}{63} = 0 \Rightarrow 16 x + 63 y + 195 = 0$

Now, the internal centre of similitude $P' (x_{2}, y_{2}) divides C C'$ internally in the ratio $r : r' = 3 : 4$

$∴ P = (\frac{3 (8) + 4 (0)}{3 + 4}, \frac{3 (- 1) + 4 (0)}{3 + 4}) = (\frac{24}{7}, \frac{- 3}{7})$

Let $m$ be the slope of transverse common tangent and the equation of tangent is $(y + \frac{3}{7}) = m (x - \frac{24}{7})$

$\Rightarrow 7 m x - 7 y - 24 m - 3 = 0 . . . (iv)$

As we know that the shortest distance from the centre of circle to the tangent is its radius.

$\Rightarrow r = perpendicular distance from C (0, 0) to (iv)$

$\Rightarrow r = \frac{|7 m (0) + (- 7) (0) - 24 m - 3|}{\sqrt{49 m^{2} + 49}}$

$\Rightarrow 3 = \frac{|24 m + 3|}{7 \sqrt{m^{2} + 1}} \Rightarrow 3 = \frac{3}{7} (\frac{8 m + 1}{\sqrt{m^{2} + 1}})$

$\Rightarrow 49 m^{2} + 49 = 64 m^{2} + 16 m + 1 \Rightarrow 15 m^{2} + 16 m - 48 = 0$

$\Rightarrow (3 m - 4) (5 m + 12) = 0$

$∴ m = \frac{4}{3} and \frac{- 12}{5}$

When $m = \frac{4}{3}$ , the equation of the required tangent is $x \cdot \frac{28}{3} - 7 y - 24 (\frac{4}{3}) + 3 = 0 \Rightarrow 4 x - 3 y - 15 = 0$

When $m = \frac{- 12}{5}$ , the equation of the required tangent is $\frac{- 84 x}{5} - 7 y - (\frac{- 288}{5}) - 3 = 0 \Rightarrow \frac{- 84 x}{5} - 7 y + \frac{273}{0} = 0 \Rightarrow 12 x + 5 y - 39 = 0$ .

Find all common tangents of the following pairs of circles $x^{2} + y^{2} = 9 and x^{2} + y^{2} - 16 x + 2 y + 49 = 0$

Important Questions on Circle

Learn and Prepare for any exam you want !

Find all common tangents of the following pairs of circles x2+y2=9 and x2+y2-16x+2y+49=0

Important Questions on Circle

Learn and Prepare for any exam you want !

Find all common tangents of the following pairs of circles $x^{2} + y^{2} = 9 and x^{2} + y^{2} - 16 x + 2 y + 49 = 0$