Question

Two circles, each of radius $5$, have a common tangent at $\left(1,1\right)$ whose equation is $3x+4y–7=0$ then their centers are

• A. $\left(4,-5\right)$,$\left(-2,3\right)$
• B. $\left(4,-3\right)$,$\left(-2,5\right)$
• C. $\left(4,5\right)$,$\left(-2,-3\right)$
• D. None of these

Answer 1

The correct option is C

$\left(4,5\right)$,$\left(-2,-3\right)$

Explanation for the correct answer:

Let $\left(h,k\right)$ be the center of one of the circle

The line joining $\left(h,k\right)$ to $\left(1,1\right)$ will be perpendicular to the common tangent

Let $m_1$ be the slope of the tangent $3x+4y–7=0$

$\Rightarrow$ $m_1$ $=\frac{-3}{4}$

Let $m_2$ be the slope of line joining $\left(h,k\right)$ to $\left(1,1\right)$.

$\Rightarrow$ $m_2$ $=\frac{4}{3}$ $\left[\because m_1{m}_2=-1\right]$

$m_2$ is also given as $m_2=\frac{k-1}{h-1}$ $\left[\because m=\frac{y_2-y_1}{x_2-x_1}\right]$

$\Rightarrow$ $\frac{k-1}{h-1}=\frac{4}{3}$

$\Rightarrow$ $\left(k-1\right)=\frac{4}{3}\left(h-1\right)$ $...\left(i\right)$

Also the distance between center and $\left(1,1\right)$ is equal to the radius

$\Rightarrow$ ${\left(h-1\right)}^2+{\left(k-1\right)}^2=5^2$

$\Rightarrow$ ${\left(h-1\right)}^2+{\left(\frac{4}{3}\right)}^2{\left(h-1\right)}^2=5^2$ …(from $\left(i\right)$)

$\Rightarrow$ ${\left(h-1\right)}^2\left[1+\frac{16}{9}\right]=25$

$\Rightarrow$ $\frac{5}{3}\left(h-1\right)=\pm 5$

$\Rightarrow$ $\left(h-1\right)=\pm 3$

$\Rightarrow$ $h=4$ or $h=-2$

Re-substituting these values in $\left(i\right)$ we get,

$\Rightarrow$ $k-1=\frac{4}{3}\times 3$ or $k-1=\frac{4}{3}\times \left(-3\right)$

$\Rightarrow$ $k=5$ or $k=-3$

Therefore, the co-ordinates of the center are $\left(4,5\right)$ and $\left(-2,-3\right)$.

Hence, option (C) is the correct answer.