Question

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

Answer 1

Here radius of the circle $=4$

Equations of two diameters of the circle are

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

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

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

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

Substituting $x=3+2y$ in $\left(1\right)$ we get

$\Rightarrow 2(3+2y)+3y=2$

$\Rightarrow 6+4y+3y=2$

$\Rightarrow 7y=2-6=-4$

$\Rightarrow y=\frac{-4}{7}$

substitute $y=\frac{-4}{7}$ in $x=3+2y=3+2\times \frac{-4}{7}=\frac{21-8}{7}=\frac{13}{7}$

Hence centre of the circle is $(\frac{13}{7},\frac{-4}{7})$

Now the equation of the required circle is

${(x+\frac{13}{7})}^2+{(y-\frac{4}{7})}^2=4^2$

$\Rightarrow {(7x+13)}^2+{(7y-4)}^2=16\times 49$

$\Rightarrow 49x^2+169+182x+49y^2+16-56y-784=0$

$\Rightarrow 49x^2+49y^2+182x-56y+169+16-784=0$

$\Rightarrow 49x^2+49y^2+182x-56y-599=0$