Question

If the equations of two diameters of a circle are 2x + y = 6 and 3x + 2y = 4 and the radius is 10, find the equation of the circle.

Answer 1

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

The intersection point of 2x + y = 6 and 3x + 2y = 4 is (8, −10).

The diameters of a circle intersect at the centre.
Thus, the coordinates of the centre are (8, −10).
∴ h = 8, k = −10

Thus, the equation of the required circle is ${\left(x-8\right)}^2+{\left(y+10\right)}^2=a^2$ ...(1)

Also, a = 10

Substituting the value of a in equation (1):
${\left(x-8\right)}^2+{\left(y+10\right)}^2=100$
$$\begin{gathered} \Rightarrow x^2+y^2-16x+64+100+20y=100 \\ \Rightarrow x^2+y^2-16x+20y+64=0 \end{gathered}$$

Hence, the required equation of the circle is $x^2+y^2-16x+20y+64=0$.