Circles
Circle
star_batch_jee_advanced_2025
Grade 11
Question:
A circle passes through the point $(3, 4)$ and cuts the circle $x^2 + y^2 = a^2$ orthogonally. The locus of its centre is a straight line. If the distance of the straight line from the origin is $817$, then find the value of $a^2 - 8140$.
Step-by-Step Solution
Key Concept: Two circles intersect orthogonally when $2g_1g_2 + 2f_1f_2 = c_1 + c_2$; use this condition along with the point constraint to find the locus.
Let the circle equation be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since it passes through $(3,4)$: $6g + 8f + c = -25$. For orthogonal intersection with $x^2 + y^2 = a^2$, we need $2g(0) + 2f(0) = c - a^2$, giving $c = a^2$. From $6g + 8f + a^2 = -25$, the locus of center $(-g, -f)$ is $6x + 8y - (a^2 + 25) = 0$. The distance from origin is $\frac{a^2 + 25}{\sqrt{100}} = \frac{a^2 + 25}{10} = 817$, yielding $a^2 = 8145$.
Correct Answer: 5