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A tangent to the circle $x^2+y^2=5$ at the point $(1,-2) \ldots .$. the circle $x^2+y^2-8 x+6 y+20=0$
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equation of tangent to the circle $x^2+y^2=5$ at $\left(x_1, y_1\right)$will be $\mathrm{xx}_1+\mathrm{yy}_1=5$
$\begin{aligned}& x \times 1+y(-2)=5 \\& x-2 y=5 \text { or } x=(2 y+5)\end{aligned}$
now this is tangent at
circle $x^2+y^2-8 x+6 y+20=0$
or $(2 y+5)^2+y^2-8(2 y+5)+6 y+20=0$
or $4 y^2+20 y+25+y^2-16 y-40+6 y+20=0$
or $5 y^2+10 y+5=0$
or $\mathrm{y}^2+2 \mathrm{y}+1=0$
$(y+1)^2=0$
$\mathrm{y}=-1$
$\therefore \mathrm{x}=2(-1)+5=3$
$\therefore$ point at which tangent is $(3,-1)$
$\begin{aligned}& x \times 1+y(-2)=5 \\& x-2 y=5 \text { or } x=(2 y+5)\end{aligned}$
now this is tangent at
circle $x^2+y^2-8 x+6 y+20=0$
or $(2 y+5)^2+y^2-8(2 y+5)+6 y+20=0$
or $4 y^2+20 y+25+y^2-16 y-40+6 y+20=0$
or $5 y^2+10 y+5=0$
or $\mathrm{y}^2+2 \mathrm{y}+1=0$
$(y+1)^2=0$
$\mathrm{y}=-1$
$\therefore \mathrm{x}=2(-1)+5=3$
$\therefore$ point at which tangent is $(3,-1)$
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