IPMAT Indore 2026 · Quantitative Ability · Question 19

Step 1 — set coordinates. Put A=\left( 0,0 \right), B=\left( 72,0 \right), C=\left( 72,30 \right), D=\left( 0,30 \right). Then P lies on AB with AP=56, so P=\left( 56,0 \right).

The circle passes through A\left( 0,0 \right), P\left( 56,0 \right) and C\left( 72,30 \right).

Step 2 — use the general circle equation. Let

{x}^{2}+{y}^{2}+Dx+Ey+F=0

Through A\left( 0,0 \right): F=0.

Through P\left( 56,0 \right): {56}^{2}+56D=0\Rightarrow D=-56.

Through C\left( 72,30 \right):

5184+900-56\left( 72 \right)+30E=0 \Rightarrow 6084-4032+30E=0 \Rightarrow E=-\frac{342}{5}

Step 3 — radius formula.

r=1/2\sqrt{{D}^{2}+{E}^{2}-4F}=1/2\sqrt{3136+\frac{116964}{25}}=1/2\sqrt{\frac{78400+116964}{25}}

=1/2\cdot \frac{\sqrt{195364}}{5}=1/2\cdot \frac{442}{5}=\frac{221}{5}

Answer: Option 3 — \frac{221}{5}

Shortcut. Because A and P both sit on the x-axis, the centre lies on the vertical line x=28 (their perpendicular bisector). Then one distance equation in the remaining unknown finishes it — faster than the general-equation route, and it avoids the big square root.