Circles
Circle
star_batch_jee_advanced_2025
Grade 11
Question:
Two circles have centres at $(a,0)$ and $(-a,0)$ and radii $r_1$ and $r_2$ $(a > r_1 > r_2)$. Then the points of contact of the common tangents to two circles lies on the
x^2 + y^2 = a^2 + r_1r_2
x^2 + y^2 = a^2 - r_1r_2
x^2 + y^2 = a^2 - 2r_1r_2
x^2 + y^2 = a^2 + 2r_1r_2
Step-by-Step Solution
Key Concept: The locus of contact points between a tangent from one circle to another forms a circle whose equation can be derived using tangency conditions.
Given circles $S_1: (x-a)^2 + y^2 = n^2$ and $S_2: (x+a)^2 + y^2 = r_2^2$, a point on $S_1$ is $(a+n\cos\theta, n\sin\theta)$. The tangent to $S_1$ at this point is $x\cos\theta + y\sin\theta = n + a\cos\theta$. This line touches $S_2$ when $|n + a\cos\theta + a\cos\theta| = r_2$, giving $2a\cos\theta + n = \pm r_2$. If contact point is $(h,k)$ with $h = a + n\cos\theta$ and $k = n\sin\theta$, then $h^2 + k^2 = a^2 + n r_2$ after substitution and simplification. Thus $(h,k)$ lies on circle $x^2 + y^2 = a^2 + nr_2$.
Correct Answer: 1,2