ParabolaeasyPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Max Distance from Circle to Parabola Vertex = 12 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let PP be a moving point on the circle x2+y2−6x−8y+21=0x^2+y^2-6x-8y+21=0. Then the maximum distance of PP from the vertex of the parabola x2+6x+y+13=0x^2+6x+y+13=0 is
A88
B1010
C1212correct
D99
Solution
Step 1: Circle x2+y2−6x−8y+21=0x^2+y^2-6x-8y+21=0: g=−3, f=−4, c=21g=-3,\ f=-4,\ c=21, centre C(3,4)C(3,4),
r=9+16−21=4=2.r=\sqrt{9+16-21}=\sqrt4=2.
 Step 2: x2+6x+y+13=0⇒(x+3)2−9+y+13=0⇒(x+3)2=−(y+4)x^2+6x+y+13=0\Rightarrow(x+3)^2-9+y+13=0\Rightarrow(x+3)^2=-(y+4), vertex A(−3,−4)A(-3,-4). Step 3: Max distance =AC+r=AC+r:
AC=(3+3)2+(4+4)2=36+64=10.AC=\sqrt{(3+3)^2+(4+4)^2}=\sqrt{36+64}=10.
∴ AC+r=10+2=12.\therefore\ AC+r=10+2=12.
 Correct answer: (3)
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