A circle with centre (2, 3) and radius 4 intersects the line x+y=3 at the points P and Q. If the tangents at P and Q intersect at the point S(α,β), then 4α-7β is equal to ___________.

Solution

Maths sir · 5 steps

PQ is the chord of contact from S, so write T = 0 at (α, β) and match it with x + y = 3.

1. Circle equation. (x-2)² + (y-3)² = 16 expands to x² + y² - 4x - 6y - 3 = 0. The line cuts it because the distance from (2, 3) to the line is 2/√2 = √2 < 4.
2. Chord of contact. The tangents at P and Q meet at S, so PQ is the chord of contact from S(α, β). Its equation is T = 0: α x + β y - 2(x + α) - 3(y + β) - 3 = 0.

   (α - 2)x + (β - 3)y - (2α + 3β + 3) = 0

3. Match with the line. This must be the same line as x + y - 3 = 0, so (α - 2)/1 = (β - 3)/1 = (2α + 3β + 3)/3 = k. Then α = k + 2, β = k + 3.
4. Find k. Substitute: 2(k+2) + 3(k+3) + 3 = 3k, so 5k + 16 = 3k and k = -8. This gives α = -6, β = -5.
5. So. 4α - 7β = -24 + 35 = 11.

The common slip: -7. Matching the constant term with +3 instead of -3 gives k = -2 and S = (0, 1), so the answer comes out as -7, not 11.

The idea to keep. Whenever tangents at the ends of a chord meet at a point, that chord is the chord of contact T = 0 of the point. Compare it with the given line, coefficient by coefficient.

Answer from the NTA final answer key. Solution by PrepWiser.