JEEAdvanced

From the external point P(7,1) the two tangents are drawn to the…

Question

From the external point P(7,1) the two tangents are drawn to the circle x^2 + y^2 = 25, touching it at points A and B. If the area of triangle PAB equals m/n in lowest terms, find m+n.

✓ Verified answer: 27checked by our engine — not a guess

Step-by-step solution

The chord of contact of tangents from P(7,1) to x^2 + y^2 = 25 is T = 0, i.e. 7x + 1*y = 25, so 7x + y = 25.
Find A and B as the intersection of this chord with the circle. From y = 25 - 7x:
x^2 + (25-7x)^2 = 25 => x^2 + 625 - 350x + 49x^2 = 25 => 50x^2 - 350x + 600 = 0 => x^2 - 7x + 12 = 0 => x = 3 or x = 4.
So A = (3,4) and B = (4,-3).

Area of triangle PAB with P(7,1), A(3,4), B(4,-3):

Area = (1/2)| (Ax-Px)(By-Py) - (Bx-Px)(Ay-Py) |
= (1/2)| (3-7)(-3-1) - (4-7)(4-1) | = (1/2)| (-4)(-4) - (-3)(3) | = (1/2)|16 + 9| = 25/2.
m/n = 25/2, m+n = 27.

Final answer27

Stuck on a problem like this?

Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.

Solve my doubt →

More Coordinate Geometry solutions