Question

The tangent to the circle $x^2+y^2=5$ at the point $(1,-2)$ also touches the circle $x^2+y^2-8x+6y+20=0$ at

• A. $(-2,1)$
• B. $(-3,0)$
• C. $(-1,-1)$
• D. $(3,-1)$

Answer 1

The correct option is D $(3,-1)$

equation of tangent to the

circle $x^2+y^2=5$ at $(x_1,y_1)$

will be $x{x}_1+y{y}_1=5$

$x\times 1+y(-2)=5$

$x-2y=5$ or $x=(2y+5)$

now this is tangent at

circle $x^2+y^2-8x+6y+20=0$

or $(2y+5{)}^2+y^2-8(2y+5)+6y+20=0$

or $4y^2+20y+25+y^2-16y-40+6y+20=0$

or $5y^2+10y+5=0$

or $y^2+2y+1=0$

$(y+1{)}^2=0$

$y=-1$

$\therefore x=2(-1)+5=3$

$\therefore$ point at which tangent

is (3,-1)

$\star$ wrong option