Class JEE Mathematics ALL Q #1159
COMPETENCY BASED
APPLY
4 Marks 2026 JEE Main 2026 (Online) 21st January Morning Shift MCQ SINGLE
The locus of the point of intersection of tangents drawn to the circle $(x-2)^{2}+(y-3)^{2}=16$ which subtends an angle of $120^{\circ}$ is:
(A) $3x^{2}+3y^{2}-12x-18y-25=0$
(B) $x^{2}+y^{2}-12x-18y-25=0$
(C) $3x^{2}+3y^{2}+12x+18y-25=0$
(D) $x^{2}+y^{2}+12x+18y-25=0$
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Step-by-Step Solution

  1. Let the point of intersection of the tangents be $P(h, k)$.
  2. The equation of the circle is $(x-2)^{2}+(y-3)^{2}=16$. The center of the circle is $C(2, 3)$ and the radius is $r=4$.
  3. The tangents from $P(h, k)$ subtend an angle of $120^{\circ}$ at the center. Therefore, the angle between the tangent and the line joining the center to the point $P$ is $60^{\circ}$.
  4. Consider the right-angled triangle formed by the center $C$, the point $P$, and the point of tangency $T$. Then $\angle TCP = 60^{\circ}$.
  5. We have $\sin 60^{\circ} = \frac{CT}{CP} = \frac{r}{CP}$. So, $\frac{\sqrt{3}}{2} = \frac{4}{\sqrt{(h-2)^{2}+(k-3)^{2}}}$.
  6. Squaring both sides, we get $\frac{3}{4} = \frac{16}{(h-2)^{2}+(k-3)^{2}}$.
  7. Therefore, $(h-2)^{2}+(k-3)^{2} = \frac{64}{3}$.
  8. Replacing $(h, k)$ with $(x, y)$, we get $(x-2)^{2}+(y-3)^{2} = \frac{64}{3}$.
  9. Expanding, we have $x^{2}-4x+4+y^{2}-6y+9 = \frac{64}{3}$.
  10. Multiplying by 3, we get $3x^{2}-12x+12+3y^{2}-18y+27 = 64$.
  11. So, $3x^{2}+3y^{2}-12x-18y+39-64 = 0$.
  12. Thus, $3x^{2}+3y^{2}-12x-18y-25 = 0$.

Correct Answer: $3x^{2}+3y^{2}-12x-18y-25=0$

AI Suggestion: Option A

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the concepts of circles, tangents, and trigonometry to derive the locus of a point.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a series of steps involving geometric and algebraic manipulations to arrive at the final equation of the locus. This involves knowing the procedures for finding tangents, applying trigonometric relationships, and simplifying equations.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It is not a straight textbook question, but requires application of multiple concepts to solve a non-standard problem.

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