Question

Two circles each of radius $5$ units touch each other at $(1,2)$. If the equation of their common tangent is $4x+3y=10$, then the centres of the two circles, respectively, are

• A. $(3,4),(-1,10)$
• B. $(5,7),(-3,-3)$
• C. $(5,5),(-3,-1)$
• D. $(5,-3),(-3,7)$

Answer 1

The correct option is C $(5,5),(-3,-1)$

Any circle which touches the line $4x+3y=10$ at $(1,2)$ is
$(x-1{)}^2+(y-2{)}^2+\lambda (4x+3y-10)=0$
i.e. $x^2+y^2+(4\lambda -2)x+(3\lambda -4)y+5-10\lambda =0$
its centre is $(1-2\lambda ,2-\frac{3\lambda }{2})$ and
radius = $\sqrt{{(1-2\lambda )}^2+{(2-\frac{3\lambda }{2})}^2-5+10\lambda }=5$
$\therefore 1+4{\lambda }^2-4\lambda +4+\frac{9{\lambda }^2}{4}-6\lambda -5+10\lambda =25$
$\frac{25{\lambda }^2}{4}=25\lambda =\pm 2$
$\therefore$ The centres are $(-3,-1),(5,5)$