Let the circle x²+y²=2ax+2by intersect the x-axis at point A ( α ,0 ) and y-axis at point B ( 0,β ) , where α β ≠ 0 . If the point C ( p,q ) lies on the chord AB, then (p+α)/(a)+(q+β)/(b) equals

IPMAT Indore 2026 · Quantitative Ability · Question 42

IPMAT Indore 2026 previous year question on Cicles, with the answer and a step-by-step solution.

CiclesCoordinate GeometryMultiple choiceMedium

Let the circle {x}^{2}+{y}^{2}=2ax+2by intersect the x-axis at point A\left( \alpha ,0 \right) and y-axis at point B\left( 0,\beta \right), where \alpha \beta \ne 0. If the point C\left( p,q \right) lies on the chord AB, then \frac{p+\alpha}{a}+\frac{q+\beta}{b} equals

Pick an option, then check.

Answer & solution

Answer: Option D

Step 1 — find the intercepts. Write the circle as {x}^{2}+{y}^{2}-2ax-2by=0 (it passes through the origin).

On the x-axis (y=0): {x}^{2}-2ax=0\Rightarrow x=0 or x=2a, so \alpha =2a.

On the y-axis (x=0): {y}^{2}-2by=0\Rightarrow y=0 or y=2b, so \beta =2b.

Step 2 — write the chord AB. It joins \left( 2a,0 \right) and \left( 0,2b \right), so in intercept form:

\frac{x}{2a}+\frac{y}{2b}=1

Step 3 — use the fact that C\left( p,q \right) lies on it.

\frac{p}{2a}+\frac{q}{2b}=1 \Rightarrow \frac{p}{a}+\frac{q}{b}=2

Step 4 — assemble the required expression.

\frac{p+\alpha}{a}+\frac{q+\beta}{b}=\frac{p+2a}{a}+\frac{q+2b}{b}=\left( \frac{p}{a}+\frac{q}{b}\right)+2+2=2+4=6

Answer: Option 4 — 6

Shortcut. Everything collapses once you see \alpha =2a and \beta =2b. Then \frac{\alpha}{a}+\frac{\beta}{b}=4 outright, and the chord condition supplies the remaining 2. No coordinates for C are ever needed.

IPMAT Indore 2026 · Quantitative Ability · Question 42

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