Find the equation of the circle which passes | vedclass

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

Question

(C) Let $S_1 = x^2+y^2-8x-6y+21=0$ and $S_2 = x^2+y^2-2x-15=0$. Using the family of circles equation $S_1 + \lambda S_2 = 0$,we have: $(x^2+y^2-8x-6y+

Vedclass · Accepted Answer

$3(x^2+y^2)-18x-12y+27=0$

Vedclass · Answer

(C) Let $S_1 = x^2+y^2-8x-6y+21=0$ and $S_2 = x^2+y^2-2x-15=0$. Using the family of circles equation $S_1 + \lambda S_2 = 0$,we have: $(x^2+y^2-8x-6y+21) + \lambda(x^2+y^2-2x-15) = 0$. Since the circle passes through the point $(1, 2)$,substitute $x=1$ and $y=2$ into the equation: $(1^2+2^2-8(1)-6(2)+21) + \lambda(1^2+2^2-2(1)-15) = 0$. $(1+4-8-12+21) + \lambda(1+4-2-15) = 0$. $6 + \lambda(-12) = 0$. $12\lambda = 6 \Rightarrow \lambda = \frac{1}{2}$. Substituting $\lambda = \frac{1}{2}$ back into the family equation: $(x^2+y^2-8x-6y+21) + \frac{1}{2}(x^2+y^2-2x-15) = 0$. Multiply the entire equation by $2$: $2(x^2+y^2-8x-6y+21) + (x^2+y^2-2x-15) = 0$. $2x^2+2y^2-16x-12y+42 + x^2+y^2-2x-15 = 0$. $3x^2+3y^2-18x-12y+27 = 0$. $3(x^2+y^2)-18x-12y+27 = 0$.

Find the equation of the circle which passes through the point $(1, 2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$.

Similar Questions

If $x-4=0$ is the radical axis of two orthogonal circles,out of which one is $x^2+y^2=36$,then the centre of the other circle is

If the circles $x^{2}+y^{2}+2x+2ky+6=0$ and $x^{2}+y^{2}+2ky+k=0$ intersect orthogonally, then $k$ is equal to

If one of the two circles $x^2+y^2+\alpha_1(x-y)+c=0$ and $x^2+y^2+\alpha_2(x-y)+c=0$ lies within the other,then (where $\alpha_1, \alpha_2 \in R, \alpha_1 \neq \alpha_2$):

If $d$ is the distance between the centres of two circles,$r_1$ and $r_2$ are their radii,and $d = r_1 + r_2$,then

$A$ circle $C$ touches the line $x=2y$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2y-5=0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1}$. Then the diameter of $C$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module