Question

If $x+2y=3$ and $2x+y=3$ touches a circle whose centre lies on a line $3x+4y=5,$ then possible centres of circle are

• A. $(\frac{5}{3},0),(3,-1)$
• B. $(\frac{5}{7},\frac{5}{6}),(4,-\frac{3}{4})$
• C. $(\frac{5}{3},0),(4,-\frac{3}{4})$
• D. $(\frac{5}{7},\frac{5}{7}),(3,-1)$

Answer 1

The correct option is D $(\frac{5}{7},\frac{5}{7}),(3,-1)$

$\because x+2y-3=0$ and $2x+y-3=0$ are non-parallel.
Bisector of lines
$\frac{(x+2y-3)}{\sqrt{5}}=\pm \frac{(2x+y-3)}{\sqrt{5}}$
$\Rightarrow (x+2y-3)=(2x+y-3)$
$\Rightarrow x-y=0$ ............................$\left(1\right)$
and $\Rightarrow (x+2y-3)=-(2x+y-3)$
$\Rightarrow x+y=2$ ............................$\left(2\right)$
$\because$ Centre of circle lies on angle bisector of lines.
Also centre lies on a line $3x+4y=5$ ..................$\left(3\right)$
Solving $\left(1\right)$ and $\left(3\right)$ we get
$(\frac{5}{7},\frac{5}{7})$
Solving $\left(2\right)$ and $\left(3\right)$ we get
$(3,-1)$
Hence the two possible centres of the circle are $(\frac{5}{7},\frac{5}{7})$, $(3,-1)$.