Question

Find the equation of the tangents through $(7,1)$ to the circle $x^2+y^2=25$.

Answer 1

$x^2+y^2=5^2$

Equation of tangent to a circle $x^2+y^2=a^2$ is $y=mx+a\sqrt{1+m^2}$

Here $a=5$

$y=mx+5\sqrt{1+m^2}$

it passes through $(7,1)$

therefore,

$1=7m+5\sqrt{1+m^2}$

$\Rightarrow {(1-7m)}^2={\left(5\sqrt{1+m^2}\right)}^2$

$\Rightarrow 1+49m^2-14m=25(1+m^2)$

$\Rightarrow {49m}^2-14m+1=25+25m^2$

${24m}^2-14m-24=0$

$\Rightarrow {12m}^2-7m-12=0$

$m=\frac{4}{3}$ and $\frac{-3}{4}$

$\therefore$ required tangents are

$(y-1)=\frac{4}{3}(x-7)$ and $(y-1)(-3/4)(x-7)$

or $4x-3y-25=0$ and $3x+4y-25=0$