Question

If the equations of the two diameters of a circle are $2x+y=1$ and $x-2y=4$ and the radius of the circle is $5$, find the equation of the circle.

Answer 1

Here radius of the circle $=5$

Equations of two diameters of the circle are

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

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

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

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

Substituting $x=4+2y$ in $\left(1\right)$ we get

$2(4+2y)+y=1$

$\Rightarrow 8+4y+y=1$

$\Rightarrow 5y=1-8=-7$

$\therefore y=\frac{-7}{5}$

substitute $y=\frac{-7}{5}$ in $x=4+2y$

we get

$x=4+2\times \frac{-7}{5}=4-\frac{14}{5}=\frac{20-14}{5}=\frac{6}{5}$

Hence centre of the circle is $(\frac{6}{5},\frac{-7}{5})$

Now the equation of the required circle is

${(x-\frac{6}{5})}^2+{(y+\frac{-7}{5})}^2=5^2$

$\Rightarrow \frac{1}{25}[{(5x-6)}^2+{(5y-7)}^2]=25$

$\Rightarrow 25x^2+36-60x+25y^2+49-70y=625$

$\Rightarrow 25x^2+25y^2-60x-70y-540=0$

or $5x^2+5y^2-12x-14y-108=0$