Question

The point/ points of intersection of the common tangents of the two circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2y-15=0$ is/are

• A. $(5,8),(-4,3)$
• B. $(8,5)$
• C. $(3,1)$
• D. $(2,1),(4,3)$

Answer 1

Answer: (B)

Given circle,
$s_1:x^2+y^2-8x-6y+21=0$
and $S_2:x^2+y^2-2y-15=0$
Circle $S_1$ : Centre $C_1(4,3)$, radius $r_1=\sqrt{16+9-21}=2$
$S_2$ : Centre $C_2(0,1)$, radius $r_2=\sqrt{1+15}=4$
Now $C_1{C}_2=\sqrt{4^2+2^2}=\sqrt{20}$
$\therefore C_1{C}_2<r_1+r_2$
$\Rightarrow$ circle $C_1$ and $C_2$ put with other.

Let point of intersection of tangent on the circle $S_1$ and $S_2$ is $P$.
Now, point $P$ divide line joining $C_2{C}_1$ in the ratio of their radii externally $\therefore \frac{C_{2}P}{C_{1}P}=\frac{r_2}{r_1}$
$\Rightarrow \frac{C_{2}P}{C_{1}P}=\frac{4}{2}=\frac{2}{1}$
$\therefore$ By using external division formula
$=\left(\frac{m_1{n}_2-m_2{n}_1}{m_1-m_2},\frac{m_1{y}_2-m_2{y}_1}{m_1-m_2}\right)$
$=\left(\frac{2\times 4-1\times 0}{2-1},\frac{2\times 3-1\times 1}{2-1}\right)=(8,5)$