Question
The circle passing through the intersection of the circles, $x^{2}+y^{2}-6 x=0$ and $x^{2}+y^{2}-4 y=0$, having its centre on the line, $2 x-3 y+12=0$, also passes through the point: (a) $(-1,3)$(b) $(-3,6)$(c) $(-3,1)$ (d) $(1,-3)$
Step 1
So, we have \[S_1 + \lambda S_2 = x^2 + y^2 - 6x + \lambda(x^2 + y^2 - 4y) = 0.\] This simplifies to \[(1 + \lambda)x^2 + (1 + \lambda)y^2 - 6x - 4\lambda y = 0.\] The center of this circle is given by $(g, f) = \left(\frac{3}{1 + \lambda}, \frac{2\lambda}{1 + Show more…
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