Question

Consider, S₁ : x(x – 6) + y(y – 8) = 0 and L : 4x + 3y = 24. S₂ is another circle of maximum radius which lies completely in 1^st quadrant and touches the line L and circle S₁.

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Linked Question 1

If centre of the S₂ is (a, b) then (a + 2b) is equal to

(i) Radius of circle S₂ is $\frac{5}{2}$

$a = 3 + \frac{5}{2} \cdot \frac{4}{5} = 5 (∵ tanθ = \frac{3}{4})$

$b = 4 + \frac{5}{2} \cdot \frac{3}{5} = \frac{11}{2}$

$∴$ a + 2b = 5 + 11 = 16

Linked Question 2

If P is a variable point on S₁ and PA and PB are pair of tangents to the circle S₂ then locus of circumcentre of $∆$ PAB is

${(x - 4)}^{2} + {(y - \frac{11}{2})}^{2} = \frac{5}{4}$

${(x - 4)}^{2} + {(y - \frac{19}{4})}^{2} = \frac{25}{4}$

${(x - 5)}^{2} + {(y - \frac{19}{4})}^{2} = \frac{25}{16}$

${(x - 4)}^{2} + {(y - \frac{11}{2})}^{2} = \frac{25}{4}$

(ii) P $\equiv$ (3 + 5 cos θ, 4 + 5 sin θ)

Let circumcentre of $∆$ PAB be (h, k)

$h = \frac{3 + 5 cosθ + 5}{2} = 4 + \frac{5}{2} cosθ$

$k = \frac{4 + 5 sinθ + \frac{11}{2}}{2} = \frac{19}{4} + \frac{5}{2} sinθ$

$∴$ Locus of (h,k) is (x – 4)² + ${(y - \frac{19}{4})}^{2} = \frac{25}{4} .$