Question

If the equation of the circle passing through the point $(8, 8)$ and having the lines $x + 2y - 2 = 0$ and $2x + 3y - 1 = 0$ as its diameters is $x^2 + y^2 + px + qy + r = 0$, then $p^2 + q^2 + r =$

• A. $244$
• B. $100$
• C. $-44$
• D. $44$

Hint:

The center of a circle is the intersection of its diameters. Use a point on the circle to find the radius, then form the equation and compute required expressions.

Answer 1

The Correct Option is C

The center of the circle is the intersection of the diameters $x + 2y - 2 = 0$ and $2x + 3y - 1 = 0$. Solve: \[ x + 2y = 2 \quad (1) \] \[ 2x + 3y = 1 \quad (2) \] Multiply (1) by 2: \[ 2x + 4y = 4 \] Subtract (2): \[ (2x + 4y) - (2x + 3y) = 4 - 1 \implies y = 3 \] Substitute $y = 3$ into (1): \[ x + 2 \cdot 3 = 2 \implies x + 6 = 2 \implies x = -4 \] Center is $(-4, 3)$. The circle’s equation is: \[ (x + 4)^2 + (y - 3)^2 = r^2 \] Since it passes through $(8, 8)$: \[ (8 + 4)^2 + (8 - 3)^2 = r^2 \implies 12^2 + 5^2 = 144 + 25 = 169 = r^2 \] Expand the circle equation: \[ (x^2 + 8x + 16) + (y^2 - 6y + 9) = 169 \] \[ x^2 + y^2 + 8x - 6y + 25 - 169 = 0 \implies x^2 + y^2 + 8x - 6y - 144 = 0 \] Thus, $p = 8$, $q = -6$, $r = -144$. Compute: \[ p^2 + q^2 + r = 8^2 + (-6)^2 + (-144) = 64 + 36 - 144 = -44 \] Option (3) is correct. Options (1), (2), and (4) do not match.