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  • Radical axis of both the circles is s1−s2=x+y=0s_{1}-s_{2}=x+y=0s1s2=x+y=0
    Taking any point on the radical axis as center, a unique circle with radius as length of tangent to any of the given two circles can be drawn which is orthogonal to both the circles. Let any point on x+y=0x+y=0x+y=0 be (λ,−λ),(\lambda,-\lambda),(λ,λ), so the radius of the circle =λ2+λ2+6λ+5=\sqrt{\lambda^{2}+\lambda^{2}+6 \lambda+5}=λ2+λ2+6λ+5
    So, the circle cutting both circles orthogonally will be (x−λ)2+(y+λ)2=2λ2+6λ+5(x-\lambda)^{2}+(y+\lambda)^{2}=2 \lambda^{2}+6 \lambda+5(xλ)2+(y+λ)2=2λ2+6λ+5
    l.e., (x2+y2−5)−2λ(x−y+3)=0\left(x^{2}+y^{2}-5\right)-2 \lambda(x-y+3)=0(x2+y25)2λ(xy+3)=0
    ∴\therefore All such circles pass through the points of intersection of x2+y2=5x^{2}+y^{2}=5x2+y2=5 and x−y+3=0x-y+3=0xy+3=0
    So, all circles pass through the two fixed points, viz (-1,2) and (-2,1)

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