AP EAMCET · Maths · Circle

Given two fixed points $A (- 2, 1)$ and $B (3, 0)$ , find the locus of a point $P$ which moves such that the angle $∠ A P B$ is always a right angle.

A $x^{2} + y^{2} + x + y + 6 = 0$
B $x^{2} + y^{2} - x - y - 6 = 0$
C $x + y + 6 = 0$
D $2 x^{2} + 2 y^{2} - 2 x - 2 y + 1 = 0$

Verified Solution

Answer & Solution

Correct Answer

(B) $x^{2} + y^{2} - x - y - 6 = 0$

Step-by-step Solution

Detailed explanation

Let, the coordinates of point Ph, k Given, points A-2, 1 and B3, 0, So, the slope of AP is (Say m1) k-1h+2, and, the slope of BP is (Say m2 )k-0h-3=kh-3 As, ∠APB=90° Hence, m1m2=-1…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Given two fixed points A(-2,1) and B(3,0), find the locus of a point P which moves such that the angle ∠APB is always a right angle.

Step-by-step Solution

Given two fixed points $A (- 2, 1)$ and $B (3, 0)$ , find the locus of a point $P$ which moves such that the angle $∠ A P B$ is always a right angle.