Question

If the equations of the two diameters of a circle are $x+2y=3$ and $x-y=4$ and the radius of the circle is $6$, find the equation of the circle.

Answer 1

Here radius of the circle $=6$

Equations of two diameters of the circle are

$x+2y=3$ ....$\left(1\right)$

$x-y=4$ ....$\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ we get

$\left(2\right)\Rightarrow x=4+y$

Substituting $x=4+y$ in $\left(1\right)$ we get

$4+y+2y=3$

$\Rightarrow 4+3y=3$

$\Rightarrow 3y=3-4=-1$

$\therefore y=\frac{-1}{3}$

substitute $y=\frac{-1}{3}$ in $x=4+y$

we get

$x=4-\frac{1}{3}=\frac{12-1}{3}=\frac{11}{3}$

Hence centre of the circle is $(\frac{11}{3},\frac{-1}{3})$

Now the equation of the required circle is

${(x-\frac{11}{3})}^2+{(y+\frac{1}{3})}^2=6^2$

$\Rightarrow x^2+\frac{121}{9}-\frac{22x}{3}+y^2+\frac{1}{9}+\frac{2y}{3}=36$

$\Rightarrow x^2+y^2-\frac{22x}{3}+\frac{2y}{3}+\frac{121+1}{9}-36=0$

$\Rightarrow x^2+y^2-\frac{22x}{3}+\frac{2y}{3}+\frac{122-324}{9}=0$

$\Rightarrow x^2+y^2-\frac{22x}{3}+\frac{2y}{3}-\frac{202}{9}=0$

$\Rightarrow 9x^2+9y^2-66x+6y-202=0$