Question

What is the equation of tangent to the circle $x^2+y^2-4x-6y+3=0$, which makes a ${45}^{\circ }$ angle with $x$ axis?

Answer 1

Eq link . $x^2+y^2-4x-6y+3=0$ ___(1) target stops $\left(m\right)=\tan {45}^{\circ }=1$

Eq. of target passing through $p(a,b)$

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

$y-b=x-a$

$x-y+(b-a)=0$ __(2)

$\perp r$ distance from centre 'o' to 'p' = radius

$=\sqrt{4+9-3}$

$=\sqrt{10}$

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

$-1+(b-a)=\sqrt{20}$

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

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

$\therefore$ Equation of tangent $x-y+(1\pm 2\sqrt{5})=0$

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

Required tangents