Question

If the circles $2x^2+2y^2+px+6y-10=0$ and $3x^2+3y^2+15x+py+21=0$ are orthogonal, then the value of $p$ is

• A. $\frac{7}{8}$
• B. $\frac{5}{8}$
• C. $\frac{8}{7}$
• D. $\frac{8}{5}$

Answer 1

Answer: (C)

$\frac{8}{7}$

If two circles are orthogonal , then
$r_1^2+r_2^2=d^2$
where, $r_1$ and $r_2$ are radias of circle
And $d$ is distance between circles.
$\therefore [{\left(\frac{p}{4}\right)}^2+{\left(\frac{3}{2}\right)}^2+5]+[{\left(\frac{5}{2}\right)}^2+{\left(\frac{p}{6}\right)}^2-7]={(\frac{p}{4}-\frac{5}{2})}^2+{(\frac{3}{2}-\frac{p}{6})}^2$
$-2=-\frac{5p}{4}-\frac{P}{2}$

$\Rightarrow -2=\frac{-7p}{4}$

$\Rightarrow P=\frac{8}{7}$

Hence, option C.