Find-the-equation-of-the-circle-through-the-points-of-intersection-of-x-2-y-2-1-0-x-2-y-2-2x-4y-1-0-and-touching-the-line-x-2y-0-

Question Number 135000 by bobhans last updated on 09/Mar/21

Find the equation of the circle through the points of intersection of x^2+y^2−1=0,x^2+y^2−2x−4y+1=0 and touching the line x+2y=0?

Answered by EDWIN88 last updated on 09/Mar/21

Let C is equation of circle, so

C \equiv x^{2} + y^{2} - 1 + λ (- 2 x - 4 y + 2) = 0

\equiv x^{2} + y^{2} - 2 λ x - 4 λ y + 2 λ - 1 = 0

witb center point at (λ, 2 λ) and radius

r = \sqrt{λ^{2} + 4 λ^{2} + 1 - 2 λ} = \frac{∣ λ + 4 λ ∣}{\sqrt{5}}

\Rightarrow \sqrt{5 (5 λ^{2} + 1 - 2 λ)} = ∣ 5 λ ∣

\Rightarrow 25 λ^{2} + 5 - 10 λ = 25 λ^{2}; λ = \frac{1}{2}

so we get solution \equiv x^{2} + y^{2} - x - 2 y = 0

Commented by bobhans last updated on 09/Mar/21

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