Question

If the equations of the two diameters of a circle are $x+y=6$ and $x+2y=4$ and the radius of the circle is $10$, find the equation of the circle.

Answer 1

Here radius of the circle $=10$

Equations of two diameters of the circle are

$x+y=6$ ....$\left(1\right)$

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

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

$\left(1\right)\Rightarrow y=6-x$

Substituting $y=6-x$ in $\left(2\right)$ we get

$x+2(6-x)=4$

$\Rightarrow x+12-2x=4$

$\Rightarrow -x=4-12=-8$

$\therefore x=8$

substitute $x=8$ in $y=6-x$

we get

$y=6-8=-2$

Hence centre of the circle is $(8,–2)$

Now the equation of the required circle is

${(x-8)}^2+{(y+2)}^2={10}^2$

or $x^2-16x+64+y^2+4y+4-100=0$

or $x^2+y^2-16x+4y-32=0$