Question

If $c_1:x^2+y^2-20x+64=0$ and$c_2:$ $x^2+y^2+30x+144=0$. Then the length of the shortest line segment $PQ$ which is tangent to $c_1$ at $P$ and $c_2$ at $Q$ is

• A. $20$
• B. $15$
• C. $22$
• D. $27$

Answer 1

The correct option is A $20$

$x^2+y^2-20x+64=0⟹(x-10{)}^2+y^2=6^2$

$x^2+y^2+30x+144=0⟹(x+15{)}^2+y^2=9^2$

The circles are actually as shown in the figure on the cartesian plane. The transverse common tangents are the shortest tangents and hence, we need to find their length.

Now, make constructions as shown in the figure such that $PQ||AR$ and $BR||AP$. Also, $AB=25$

$\therefore$ in $△ABR$, ${25}^2={15}^2+A{R}^2⟹AR=20$.

$\therefore PQ=AR=20$

This is the required solution.