Question

If the chord of contact of tangents from a point P to a given circle passes through Q, then the circle on PQ as diameter.

• A. cuts the given circle orthogonally
• B. touches the given circle externally
• C. touches the given circle internally
• D. none of these

Answer 1

The correct option is A cuts the given circle orthogonally

Let S:$x^2+y^2+2gx+2fy+c=0$ be the circle'

Chord of contact from P($x_1,y_1)$is given by,

$x{x}_1+y{y}_1+g(x+x_1)+f(y+y_1)+c=0$

Since $Q(x_2,y_2)$ lies on the above line

$x_1{x}_2+y_1{y}_2+g(x_1+x_2)+f(y_1+y_2)+c=0-\left(1\right)$

Centre of $S_1=(-g,-f),r_1=\sqrt{g^2+f^2-c}$

Midpoint of PQ=$C-2(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$r_2=\frac{\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}}{2}$

Circle with centre $c_2$ and radius $r_2$ is given by

$S_2:(x-(\frac{x_1+x_2}{2}){)}^2+(y-9\frac{y_1+y_2}{2}){)}^2=r_2^2$

$C_1{C}_2=\sqrt{(\frac{x_1+x_2}{2}+g{)}^2+(\frac{y_1+y_2}{2}+f{)}^2}$

Simplifying above equation usong (1)

$C_1{C}_2^2=\frac{(x_1-x_2{)}^2}{4}+\frac{(y_1-y_2{)}^2}{4}+g^2+f^2-c$

$r_1^2+r_2^2=(g^2+f^2-c)+\frac{(x_1-x_3{)}^2+(y_1-y_2{)}^2}{4}$

$⟹C_1{C}_2^2=r_1^2+r_2^2$

$S_1$ abd $S_2$ cut orthogonally.