Question

Find the equations of the tangents to the circle $x^2+y^2-6x-4y+5=0$, which make an angle of ${45}^o$ with the axis of x.

Answer 1

Given,

a circle equation as $x^2+y^2-6x-4y+5=0$

angle $\left(\theta \right)={45}^o$

$\iff \tan {45}^o=1$

Also given, to find the equation of tangents.

Let,

$x^2+y^2-6x-4y+5=0---\left(1\right)$

Now,

Equation of tangents passing through $P(a,b)$ is,

$(y-b)=m(x-a)$

$(\because m=1)$

$\Rightarrow y-b=x-a$

$\Rightarrow x-y=b-a$

$\Rightarrow x-y-b+a=0$

$x-y+(b-a)=0$

let it be eq $\left(2\right)$

here,

perpendicular distance from centre ${}^{\prime }{O}^{\prime }$ to ${}^{\prime }{P}^{\prime }$

radius $\left(r\right)=\sqrt{(2{)}^2+(3{)}^2-3}=\sqrt{4+9-3}$

$R=\sqrt{10}$.

$\Rightarrow y=mx+c\iff \therefore m=1\Rightarrow y=x+C$.

$\sqrt{10}=\frac{|1\left(2\right)-1\left(3\right)+(b-a)|}{\sqrt{1+1}}$

$\sqrt{2}.\sqrt{10}=|2-3+(b-a)|$

$\sqrt{20}=|(b-a)+2-3|$

$\pm \sqrt{4\times 5}=2+(b-a)-3$

$\pm 2\sqrt{5}=-1+(b-a)$

$(b-a)=1\pm 2\sqrt{5}---$ eq $\left(3\right)$

$\therefore$ we get equation of tangent,

put $(b-a)$ value in eq $\left(2\right)$ we get,

$x-y-[1\pm 2\sqrt{5}]=0$

$\iff x-y+(1+2\sqrt{5})=0,x-y+(1-2\sqrt{5})=0$

$\therefore$ the tangents for circle $x^2+y^2-6x-4y+5=0$

are $x-y+(1+2\sqrt{5})=0$ & $x-y+(1-2\sqrt{5})=0$