JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Circle

Let AB be a chord of length 12 of the circle

(x - 2)^{2} + (y + 1)^{2} = \frac{169}{4}

.

If tangents drawn to the circle at points A and B intersect at the point P, then five times the distance of point P from chord AB is equal to ____.

Answer: 72

Solution

The circle has center

C (2, - 1)

and radius

r = \frac{13}{2}

.

Let M be the midpoint of chord AB. Then

C M ⊥ A B

and

A M = 6

(half the chord length).

Using the Pythagorean theorem in triangle CAM:

C M = r^{2} - A M^{2} = (\frac{13}{2})^{2} - 6^{2} = \frac{169}{4} - 36 = \frac{25}{4} = \frac{5}{2}

In triangle CAP (where P is the intersection of tangents at A and B):

cos θ = \frac{A M}{C A} = \frac{6}{\frac{13}{2}} = \frac{12}{13}

Therefore:

sin θ = \frac{5}{13}

The distance from P to chord AB is:

P M = A M cot θ = 6 \cdot \frac{12}{5} = \frac{72}{5}

Therefore:

5 \cdot P M = 72

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Circle

Solution

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