JEE Main 2020 — Circle Question with Solution
From: JEE Main 2020 (Online) 7th January Evening Slot
Question
Let the tangents drawn from the origin to the circle,
x2 + y2 - 8x - 4y + 16 = 0 touch it at the points A and B. The (AB)2 is equal to :
x2 + y2 - 8x - 4y + 16 = 0 touch it at the points A and B. The (AB)2 is equal to :
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
Equation of chord of contact is
x.0 + y.0 – 4(x + 0) –2(y + 0) + 16 = 0
2x + y – 8 = 0
Length of CM = = units
AM = BM = =
Length of chord of contact (AB) =
AB2 = =
x.0 + y.0 – 4(x + 0) –2(y + 0) + 16 = 0
2x + y – 8 = 0
Length of CM = = units
AM = BM = =
Length of chord of contact (AB) =
AB2 = =
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Circle chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2020, covering the Circle chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.