Question

The locus of midpoint of the chord of contact of $x^2+y^2=2$ from the points on $3x+4y=10$ is a circle whose centre is

• A. $(\frac{4}{5},\frac{3}{5})$
• B. $(\frac{3}{5},\frac{4}{5})$
• C. $(\frac{4}{10},\frac{3}{10})$
• D. $(\frac{3}{10},\frac{4}{10})$

Answer 1

The correct option is D $(\frac{3}{10},\frac{4}{10})$

Let the midpoint be $(h,k).$
Equation of chord of contact is
$T=S_1\Rightarrow hx+ky-2=h^2+k^2-2\Rightarrow hx+ky-(h^2+k^2)=0\cdots \left(1\right)$

Consider any point $(x_1,y_1)$ on $3x+4y=10$
Equation of chord of contact is
$T=0\Rightarrow x{x}_1+y{y}_1-2=0\cdots \left(2\right)$
Comparing equation $\left(1\right)$ and $\left(2\right)$, we get
$\frac{x_1}{h}=\frac{y_1}{k}=\frac{2}{h^2+k^2}\Rightarrow x_1=\frac{2h}{h^2+k^2},y_1=\frac{2k}{h^2+k^2}$
Now putting $(x_1,y_1)$ in $3x+4y=10$, so
$6h+8k=10h^2+10k^2\Rightarrow x^2+y^2-\frac{6x}{10}-\frac{8y}{10}=0$

Hence, the centre is $(\frac{3}{10},\frac{4}{10})$