Question

The line $4x+3y-4=0$ divides the circumference of the circle centred at $(5,3),$ in the ratio $1:2.$ Then the equation of the circle is

• A. $x^2+y^2-10x-6y-66=0$
• B. $x^2+y^2-10x-6y+100=0$
• C. $x^2+y^2-10x-6y+66=0$
• D. $x^2+y^2-10x-6y-100=0$

Answer 1

The correct option is A $x^2+y^2-10x-6y-66=0$

Let line $4x+3y-4=0$ makes an angle $\alpha$ at the centre.
Then, $\alpha +2\alpha ={360}^{\circ }$
$\Rightarrow \alpha ={120}^{\circ }$
$\therefore \angle ACB={60}^{\circ }$

Perpendicular distance of $(5,3)$ to line $4x+3y-4=0$ is
$BC=\left|\frac{4\times 5+3\times 3-4}{\sqrt{25}}\right|$
$\Rightarrow BC=5$
Now, $\cos {60}^{\circ }=\frac{BC}{AC}$
$\therefore r=AC=10$
Hence, equation of the circle is
$(x-5{)}^2+(y-3{)}^2={10}^2$
or, $x^2+y^2-10x-6y-66=0$