Find the equation of the circle passing throu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of the family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$.

Vedclass · Accepted Answer

$x^2 + y^2 - 3x + 1 = 0$

Vedclass · Answer

(B) The equation of the family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$.Here,$S_1 = x^2 + y^2 - 6 = 0$ and $S_2 = x^2 + y^2 - 6x + 8 = 0$.The equation of the required circle is $(x^2 + y^2 - 6) + \lambda(x^2 + y^2 - 6x + 8) = 0$.Since the circle passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:$(1^2 + 1^2 - 6) + \lambda(1^2 + 1^2 - 6(1) + 8) = 0$$(2 - 6) + \lambda(2 - 6 + 8) = 0$$-4 + \lambda(4) = 0$$4\lambda = 4 \implies \lambda = 1$.Substituting $\lambda = 1$ back into the equation:$(x^2 + y^2 - 6) + 1(x^2 + y^2 - 6x + 8) = 0$$2x^2 + 2y^2 - 6x + 2 = 0$Dividing by $2$,we get $x^2 + y^2 - 3x + 1 = 0$.

Find the equation of the circle passing through the point $(1, 1)$ and the intersection points of the circles $x^2 + y^2 = 6$ and $x^2 + y^2 - 6x + 8 = 0$.

Similar Questions

Which of the following crops have been brought to India from the New World?

Which of the following compounds will exhibit geometrical isomerism?

What is the entropy change in $J K^{-1}$ during the melting of $27.3 \ g$ of ice at $0^{\circ} C$? (Latent heat of fusion of ice $= 330 \ J g^{-1}$)

Which of the following plants produces seeds but not flowers?

Vedclass Test Series

Exam Paper Generator

Online Exam Module