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Question

Mathematics Question on Conic sections

Two circles x2+y2=6 x^2 + y^2 = 6 and x2+y26x+8=0 x^2 + y^2 - 6x + 8 = 0 are given. Then the equation of the circle passing through their points of intersection and the point (1, 1) is

A

x2+y26x+4=0 x^2 + y^2 - 6x + 4 = 0

B

x2+y23x+1=0 x^2 + y^2 - 3x + 1 =0

C

x2+y24y+2=0x^2 + y^2 - 4y + 2 = 0

D

none of these.

Answer

x2+y23x+1=0 x^2 + y^2 - 3x + 1 =0

Explanation

Solution

The required equation of circle is, S1+λ(S2 S1)=0S_1 + \lambda (S_2 ~ S_1 ) = 0
(x2+y26)+λ(6x+14)=0\therefore \, \, \, \, \, \, \, \, \, \, \, ( x^2 + y^2 -6) + \lambda (-6x +14)=0
Also, passing through (1, 1).
4+λ(8)=0λ=12\Rightarrow \, \, \, \, \, \, \, \, -4 + \lambda (8) = 0 \Rightarrow \lambda = \frac{1}{2}
Requiredequationofcircleis\therefore \, Required equation of circle is
x2+y263x+7=0\, \, \, \, \, \, \, \, \, \, \, x^2 + y^2 - 6 - 3x +7 = 0
or x2+y23x+1=0 \, \, \, \, \, \, \, \, \, \, \, x^2 + y^2 - 3x + 1 = 0