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

Answer: (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

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

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

Midpoint of PQ $=C_2(\frac{x_1+x_2}{2}+\frac{y_1+y_2}{2})$

$r_2=\frac{\sqrt{(x_1–x2{)}^2+(y_1–y_2{)}^2}}{2}$

Circle with center $C_2$ and radius $r_2$ is given by

$S2:(x-\left(\frac{x_1+x_2}{2}\right){)}^2+(y-\left(\frac{y_1+y_2}{2}\right){)}^2=r_2^2$

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

$=\sqrt{\frac{x_1^2+x_2^2+2x_1{x}_2}{4}+g^2+g(x_1+x_2)+\frac{y_1^2+y_2^2+2y_1{y}_2}{4}+f^2+f(y_1+y_2)}$

Simplifying using (1)

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

$r_1^2+r_2^2=(g^2+f^2–c)+\frac{(x_1–x_2{)}^2+(y_1–y_2{)}^2}{4}$

$⟹C_1{C}_2^2=r_1^2+r_2^2$

$S_1$ and $S_2$ cut orthogonally.