Question
For any real $k$, the circle $x^{2} + y^{2} + 2kx + 2ky - 8 = 0$ passes through two fixed points $A$ and $B$. Locus of the point of intersection of the tangents to the circle at $A$ and $B$ is:
Ans. (c)
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Solution
[Key Concept: $x^{2} + y^{2} + 2kx + 2ky - 8 = 0$ for all $k$: group the $k$-free and $k$-coefficient parts to get the fixed points.]
Given family of circles:
x² + y² + 2kx + 2ky − 8 = 0
For two different values k₁ and k₂, the corresponding equations are:
x² + y² + 2k₁x + 2k₁y − 8 = 0
x² + y² + 2k₂x + 2k₂y − 8 = 0
Subtracting,
2(k₁ − k₂)(x + y) = 0
Since k₁ ≠ k₂,
x + y = 0
Therefore, the common chord AB of all the circles is:
AB: x + y = 0
Let P(h, l) be the point of intersection of the tangents drawn to the circle at A and B.
The circle can be written as:
(x + k)² + (y + k)² = 2k² + 8
The polar of P(h, l) with respect to this circle is:
(x + k)(h + k) + (y + k)(l + k) = 2k² + 8
Since AB is the polar of P and
AB: x + y = 0,
the coefficients of x and y must be equal.
Therefore,
h + k = l + k
Hence,
h = l
Thus, P always lies on the line:
x = y
Therefore, the locus of P is:
x = y
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