AP EAMCET · Maths · Straight Lines

The locus of a point which is at a distance of $4$ units from $(3, - 2)$ in $x y$ -plane is

A $x^{2} + y^{2} + 6 x - 4 y + 16 = 0$
B $x^{2} + y^{2} - 6 x - 4 y + 3 = 0$
C $x^{2} + y^{2} - 6 x + 4 y - 16 = 0$
D $x^{2} + y^{2} - 6 x + 4 y - 3 = 0$

Verified Solution

Answer & Solution

Correct Answer

(D) $x^{2} + y^{2} - 6 x + 4 y - 3 = 0$

Step-by-step Solution

Detailed explanation

Let the point be h,k Distance between the two points will be =h-32+k+22 ⇒4=h-32+k+22 Squaring both sides we get, 16=h2+9-6h+k2+4k+4 ⇒h2-6h+k2+4k-3=0 Therefore, required locus is x2+y2-6x+4y-3=0

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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The locus of a point which is at a distance of 4 units from (3,-2) in xy-plane is

Step-by-step Solution

The locus of a point which is at a distance of $4$ units from $(3, - 2)$ in $x y$ -plane is