First slide

Analysis of two circles in circles

Question

One of the limit point of the coaxial system of circles containing $x^{2} + y^{2} - 6 x - 6 y + 4 = 0, x^{2} + y^{2} - 2 x - 4 y + 3 = 0,$ is

Moderate

A

(-1,1)
B

(-1,2)
C

(-2,1)
D

(-2,2)

Solution

The equation of the family of coaxial system of circles haveing $x^{2} + y^{2} - 6 x - 6 y + 4 = 0 and x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
two members is
$x^{2} + y^{2} - 6 x - 6 y + 4 + λ (- 4 x - 2 y + 1) = 0$
$\Rightarrow x^{2} + y^{2} - 2 x (3 + 2 λ) - 2 y (3 + λ) + 4 + λ = 0$
Coordinates of centre of circle $(i) are (3 + 2 λ, 3 + λ) .1$
Radius $= \sqrt{(3 + 2 λ)^{2} + (3 + λ)^{2} - (4 + λ)}$
For limiting points, we must have
Radius $= 0 \Rightarrow 5 λ^{2} + 17 λ + 14 = 0 \Rightarrow λ = - 2, - 7 / 5$
Hence, limiting points are $(- 1, 1) and (1 / 5, 8 / 5)$

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