Let the centre of a circle C be (α,β) and its radius r<8. Let 3x+4y=24 and 3x−4y=32 be two tangents and 4x+3y=1 be a normal to C. Then (α−β+r) is equal to
(A)7✓ Correct
(B)9
(C)5
(D)6
Answer: (A)
Solution
Since 4x+3y=1 is a normal, the centre (α,β) lies on it, so 4α+3β=1. The centre is equidistant from the two tangents, giving 53α+4β−24=53α−4β−32. Taking 3α+4β−24=3α−4β−32 yields 8β=−8, so β=−1, and then 4α−3=1 gives α=1. The radius is r=5∣3(1)+4(−1)−24∣=525=5, which satisfies r<8. The other bisector leads to α=328 with r=980>8, which is rejected. Therefore α−β+r=1−(−1)+5=7.
JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Single correct option