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Mathematics · Circles

JEE Main 2026 — 5 April, Evening Shift — Question 35

Let the point P be the vertex of the parabola y = x² - 6x + 12. If a line passing through the point P intersects the circle x² + y² - 2x - 4y + 3 = 0 at the points R and S, then the maximum value of (PR + PS)² is:

  1. Option A:

    10

  2. Option B:

    20

    Correct
  3. Option C:

    25

  4. Option D:

    5

Answer: B

Step-by-step solution

Parabola y=x2−6x+12=(x−3)2+3y−3=(x−3)2⇒\mathrm{y}=\mathrm{x}^{2}-6 \mathrm{x}+12=(\mathrm{x}-3)^{2}+3 \mathrm{y}-3=(\mathrm{x}-3)^{2} \Rightarrow Vertex P (3,3)(3,3) x2+y2−2x−4y+3=0x^{2}+y^{2}-2 x-4 y+3=0 C(1,2);r=2\mathrm{C}(1,2) ; \mathrm{r}=\sqrt{2} For max. of (PR+PS)2(\mathrm{PR}+\mathrm{PS})^{2} RS should be diameter

(PR+PS)max⁡2=(PC−r+PC+r)2=(2PC)2=4(PC)2=4×5=20\begin{aligned} (\mathrm{PR}+\mathrm{PS})_{\max }^{2} & =(\mathrm{PC}-\mathrm{r}+\mathrm{PC}+\mathrm{r})^{2} \\& =(2 \mathrm{PC})^{2}=4(\mathrm{PC})^{2} \\& =4 \times 5=20 \end{aligned}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Mathematics
Chapter
Circles
Topic
Introduction to Circles