Question

Two circles, each of radius $5$, have a common tangent at $(1,2)$ whose equation is $4x+3y-10=0$. Then their centres are

• A. $(4,-5),(-2,3)$
• B. $(4,-3),(-2,5)$
• C. $(5,5),(-3,-1)$
• D. none of these

Answer 1

Answer: (C)

$(5,5),(-3,-1)$

Let $(\alpha ,\beta )$ be the centre of on of the circle. Then, centre must lie on the line $\perp$ to the common tangent $4x+3y=10$ and passing through the point $(1,2).$ Thus, the equation of the radius is

$3x-4y+k=0$ ...(1)

Since it passes through $(1,2),$ therefore $3-8+k=0$ i.e., $k=5$

Substituting $k=5$ in (1), the equation of the line joining centres is

$3x-4y+5=0$ ...(2)

As center lies on (2), we have $3\alpha -4\beta +5=0$

$\Rightarrow \beta =\frac{3\alpha +5}{4}$ ...(3)

Since the radius of circle is $5,$ therefore

${(\alpha -1)}^2+{(\beta -2)}^2=25$ ...(4)

Substituting the value of $\beta$ from (3) in (4), we get

${(\alpha -1)}^2+{\left(\frac{3\alpha +5}{4}\right)}^2=25$

$\Rightarrow {(\alpha -1)}^2+\frac{{(3\alpha -3)}^2}{16}=25$

$\Rightarrow {(\alpha -1)}^2+\frac{9}{16}{(\alpha -1)}^2=15$

$\Rightarrow {(\alpha -1)}^2(1+\frac{9}{16})=25\Rightarrow {(\alpha -1)}^2=16$

$\Rightarrow {(\alpha -1)}^2=\pm 4\Rightarrow \alpha =5$ or $\alpha =-5$ ...(5)

From (3)and (5), we have $\alpha =5,\beta =5$

or, $\alpha =-3,\beta =-1.$

Thus, the cerntres of the circle are $(5,5)$ and $(-3,-1)$