Question

If the equations of the two diameters of a circle are $2x-y=1$ and $3x-2y=3$ 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

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

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

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

$\left(1\right)\Rightarrow y=2x-1$

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

$\Rightarrow 3x-2(2x-1)=3$

$\Rightarrow 3x-4x+2=3$

$\Rightarrow -x=3-2=1$

$\Rightarrow x=-1$

substitute $x=-1$ in $y=2x-1=2\times -1-1=-2-1=-3$

Hence centre of the circle is $(-1,-3)$

Now the equation of the required circle is

${(x+1)}^2+{(y+3)}^2=5^2$

$\Rightarrow x^2+1+2x+y^2+6y+9-25=0$

$\Rightarrow x^2+y^2+2x+6y-15=0$