SATHEE: Circle 3 Question 21

Circle 3 Question 21

21. Find the coordinates of the point at which the circles $x^{2} - y^{2} - 4 x - 2 y + 4 = 0$ and $x^{2} + y^{2} - 12 x - 8 y + 36 = 0$ touch each other. Also, find equations of common tangents touching the circles the distinct points.

(1993, 5M)

Show Answer

Answer:

Correct Answer: 21. $(x - 5)^{2} + (y - 5)^{2} = 5^{2}$ and $(x + 3)^{2} + (y + 1)^{2} = 5^{2}$

Solution:

Two circles touch each other externally, if $C_{1} C_{2} = r_{1} + r_{2}$ and internally if $C_{1} C_{2} = r_{1} \sim r_{2}$

Given circles are $x^{2} + y^{2} - 4 x - 2 y + 4 = 0$ ,

whose centre $C_{1} (2, 1)$ and radius $r_{1} = 1$

and $x^{2} + y^{2} - 12 x - 8 y + 36 = 0$

whose centre $C_{2} (6, 4)$ and radius $r_{2} = 4$

The distance between the centres is

$\sqrt{(6 - 2)^{2} + (4 - 1)^{2}} = \sqrt{16 + 9} = 5$

$\Rightarrow C_{1} C_{2} = r_{1} + r_{2}$

Therefore, the circles touch each other externally and at the point of touching the point divides the line joining the two centres internally in the ratio of their radii, $1 : 4$ .

Therefore, $x_{1} = \frac{1 \times 6 + 4 \times 2}{1 + 4} = \frac{14}{5}$

$y_{1} = \frac{1 \times 4 + 4 \times 1}{1 + 4} = \frac{8}{5}$

Again, to determine the equations of common tangents touching the circles in distinct points, we know that, the tangents pass through a point which divides the line joining the two centres externally in the ratio of their radii, i.e. $1 : 4$ .

Therefore, $x_{2} = \frac{1 \times 6 - 4 \times 2}{1 - 4} = \frac{- 2}{- 3} = \frac{2}{3}$

and $y_{2} = \frac{1 \times 4 - 4 \times 1}{1 - 4} = 0$

Now, let $m$ be the slope of the tangent and this line passing through $(2 / 3, 0)$ is

$\begin{aligned} y - 0 & = m (x - 2 / 3) \\ \Rightarrow y - m x + \frac{2}{3} m & = 0 \end{aligned}$

This is tangent to the Ist circle, if perpendicular distance from centre $=$ radius.

$\begin{aligned} ∴ \frac{1 - 2 m + (2 / 3) m}{\sqrt{1 + m^{2}}} = 1 [∵ C_{1} \equiv (2, 1) and r_{1} = 1] \\ \Rightarrow 1 - 2 m + (2 / 3) m = \sqrt{1 + m^{2}} \\ \Rightarrow 1 - \frac{4}{3} m = \sqrt{1 + m^{2}} \\ \Rightarrow 1 + \frac{16}{9} m^{2} - \frac{8}{3} m = 1 + m^{2} \\ \Rightarrow \frac{7}{9} m^{2} - \frac{8}{3} m = 0 \\ \Rightarrow m \frac{7}{9} m - \frac{8}{3} = 0 \\ \Rightarrow m = 0, m = \frac{24}{7} \end{aligned}$