Question

Two circles each of radius $5$ units touch each other at $(1,2)$ and $4 x+3 y=10$ is their common tangent. The equation of that circle among the two given circles, such that some portion of it lies in every quadrant is

• A. $x^2 + y^2 + 6x + 2y + 15 = 0$
• B. $x^2 + y^2 + 2x + 6y - 15 = 0$
• C. $x^2 + y^2 + 6x + 2y - 15 = 0$
• D. $x^2 + y^2 - 6x + 2y - 15 = 0$

Answer 1

Answer: (C)

Given, equation of common tangent is
$4x + 3y = 10$

$4 x+3 y=10$
$\therefore $ Slope of tangent $=-\frac{4}{3}$
Slope of line perpendicular to tangent $=\frac{3}{4}$
$\therefore m=\tan \theta=\frac{3}{4}$
$\sin \theta=\frac{3}{5} $ and $ \cos \theta=\frac{4}{5}$
Now, $\frac{x-1}{\frac{4}{5}}=\frac{y-2}{\frac{3}{5}}=\pm 5$
$\Rightarrow x=\left(\pm 5 \times \frac{4}{5}+1\right) $
and $ y=\pm 5 \times \frac{3}{5}+2$
$ \Rightarrow x=(\pm 4+1)$ and $ y=(\pm 3+2) $
$\Rightarrow x=5,-3 $ and $ y=5,-1 $
So, $ C_{1}(5,5)$ and $ C_{2}(-3,-1).$
$\therefore $ Equation of required circles are
or $(x-5)^{2}+(y-5)^{2}=(5)^{2} (\because r=5)$
$(x+3)^{2}+(y+1)^{2}=(5)^{2}$
$\Rightarrow x^{2}+25-10 x+y^{2}+25-10 y=25$
or $x^{2}+9+6 x+y^{2}+1+2 y=25$
$\Rightarrow x^{2}+y^{2}-10 x-10 y+25=0 $
or $ x^{2}+y^{2}+6 x+2 y-15=0$