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

\[\left( x - 8 \right)^2 + \left( y + 10 \right)^2 = a^2\]

Also, a = 10
Substituting the value of a in equation (1):

\[\left( x - 8 \right)^2 + \left( y + 10 \right)^2 = 100\]

\[\Rightarrow x^2 + y^2 - 16x + 64 + 100 + 20y = 100\]
\[ \Rightarrow x^2 + y^2 - 16x + 20y + 64 = 0\]

Hence, the required equation of the circle is

\[x^2 + y^2 - 16x + 20y + 64 = 0\]