Question

Two circles each of radius $5$ units touch each other at $(1,2)$ and $4x+3y=10$ 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
$4x+3y=10$

$4x+3y=10$
$\therefore$ 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 $(x-5{)}^2+(y-5{)}^2=(5{)}^2(\because r=5)$
$(x+3{)}^2+(y+1{)}^2=(5{)}^2$
$\Rightarrow x^2+25-10x+y^2+25-10y=25$
or $x^2+9+6x+y^2+1+2y=25$
$\Rightarrow x^2+y^2-10x-10y+25=0$
or $x^2+y^2+6x+2y-15=0$