JEE Main 2022 · MathsNumerical answerNumerical valueMediumCalculation

JEE Main 28 June 2022, Shift 2, Maths Q23

Question 23 of 90 in this shift, Maths question 23 of 30, Section B.

If one of the diameters of the circle

x^2+y^2-2\sqrt{2}x-6\sqrt{2}y+14=0

is a chord of the circle

\left(x-2\sqrt{2}\right)^2+\left(y-2\sqrt{2}\right)^2=r^2

, then the value of

r^2

is equal to __________.

Official answer

NTA final key.

Chapter: Conic Sections
Topic: Power of a Point
Idea tested: Length of Common Chord via Radical Axis

Same idea in other shifts

Asked 3× in all

30 Jan 2024, Shift 2 · Q28Consider two circles $C_1: x^2 + y^2 = 25$ and $C_2: (x - \alpha)^2 + y^2 = 16$ , where $\alpha \in (5, 9)$ . Let the angle between the two…HardNumerical value
3 Sep 2020, Shift 2 · Q64Let the latus ractum of the parabola $y^2 = 4x$ be the common chord to the circles $C_1$ and $C_2$ each of them having radius $2\sqrt{5}$ .…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.