Question

The chord of contact of the pair of tangents drawn from any point on the line $3x+y=5$ to the circle $x^2+y^2=4$ passes through

• A. $(\frac{1}{5},\frac{2}{7})$
• B. $(\frac{12}{7},\frac{4}{7})$
• C. $(\frac{12}{5},\frac{4}{5})$
• D. $(1,1)$

Answer 1

Answer: (C)

$(\frac{12}{5},\frac{4}{5})$

$S:x^2+y^2=4$

Chord of contact from any point $P(x_1,y_1)$ is given by $T=0$

$T=x{x}_1+y{y}_1–r^2$

$3x+y=5⟹y=5–3x$

Any point on line is given by $(a,5–3a)$

$T=ax+(5–3a)y=4$

If $(x,y)=(\frac{12}{5},\frac{4}{5})$

$⟹a\times \frac{12}{5}+(5–3a)\frac{4}{5}=4$

$⟹\frac{12a}{5}+4-\frac{12a}{5}=4$

$4=4$

$⟹(\frac{12}{5},\frac{4}{5})$ lies on chord of contact.