The circle passing through the intersection of x2 + y2 - 4 = 0 and x2 + y2 - 6x + 5 = 0 which passes through the point (2.1) is
Answers
EXPLANATION.
Circle passing through the intersection.
⇒ x² + y² - 4 = 0.
⇒ x² + y² - 6x + 5 = 0.
Passes through the point = (2,1).
As we know that,
The equation of the circle passing through the points of intersection of the circle : S = 0 and L = 0 is S + λL = 0.
⇒ S₁ + λS₂ = 0.
⇒ (x² + y² - 4) = λ(x² + y² - 6x + 5).
Passes through the point = (2,1).
⇒ [(2)² + (1)² - 4] = λ[(2)² + (1)² - 6(2) + 5].
⇒ [4 + 1 - 4] = λ[4 + 1 - 12 + 5].
⇒ 1 = λ[10 - 12].
⇒ 1 = λ(-2).
⇒ λ = -1/2.
Put the value of λ = -1/2 in the equation, we get.
⇒ (x² + y² - 4) = λ(x² + y² - 6x + 5).
⇒ (x² + y² - 4) = (-1/2)(x² + y² - 6x + 5).
⇒ 2(x² + y² - 4) = (-1)(x² + y² - 6x + 5).
⇒ 2x² + 2y² - 8 = -x² - y² + 6x - 5.
⇒ 2x² + 2y² - 8 + x² + y² - 6x + 5 = 0.
⇒ 3x² + 3y² - 6x - 3 = 0.
⇒ 3(x² + y² - 2x - 1) = 0.
⇒ x² + y² - 2x - 1 = 0.
MORE INFORMATION.
(1) = Length of tangent : = √S₁.
(2) = Pair of tangents = SS₁ = T².