Question

Find the equation to the pair of tangents drawn
(1) from the point $(11,3)$ to the circle $x^2+y^2=65$,
(2) from the point $(4,5)$ to the circle $2x^2+2y^2-8x+12y+21=0$.

Answer 1

For a circle $x^2+y^2+2gx+2fy+c=0$ pair of tangents from point $(x_1,y_1)$ is given as

$(x^2+y^2+2gx+2fy+c)(x_1^2+y_1^2+2g{x}_1+2f{y}_1+c)=(x{x}_1+y{y}_1+g(x+x_1)+f(y+y_1)+c{)}^2$
Expressed as

$S{S}^{\prime }=T^2$

For circle (1) $x^2+y^2=65$

$(x^2+y^2-65)(121+9+2(11\left)\right(0)+2(0\left)\right(3)-65)=(11x+3y-65{)}^2$

$28x^2+33xy-28y^2-715x-195y+4225=0$

For circle (2) $x^2+y^2-4x+6y+\frac{21}{2}=0$

Equation of pair of tangent is

$(x^2+y^2-4x+6y+\frac{21}{2})(16+25+2(-2\left)\right(4)+2(3\left)\right(5)+\frac{21}{2})={(4x+5y-2(x+4)+3(y+5)+\frac{21}{2})}^2$

$123x^2+64xy-3y^2-664x-226y+763=0$

28x2+33xy−28y2−715x−195y+4225=0