If the circle x^2+y^2+2x+3y+1=0 cuts another | vedclass

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

Question

(B) The equations of the circles are $S_1 \equiv x^2+y^2+2x+3y+1=0$ and $S_2 \equiv x^2+y^2+4x+3y+2=0$. Since the circles intersect at $A$ and $B$,the

Vedclass · Accepted Answer

$2x^2+2y^2+2x+6y+1=0$

Vedclass · Answer

(B) The equations of the circles are $S_1 \equiv x^2+y^2+2x+3y+1=0$ and $S_2 \equiv x^2+y^2+4x+3y+2=0$. Since the circles intersect at $A$ and $B$,the equation of the common chord $AB$ is given by $S_1 - S_2 = 0$. $(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$ $-2x - 1 = 0 \Rightarrow x = -\frac{1}{2}$. The family of circles passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda(S_1 - S_2) = 0$. However,the circle with $AB$ as diameter is given by $S_1 + k(S_1 - S_2) = 0$ where the coefficient of $x^2$ and $y^2$ is $1$. Alternatively,using the property that the circle with diameter $AB$ is $S_1 + k(S_1 - S_2) = 0$,we substitute $x = -\frac{1}{2}$ into $S_1$: $S_1 = (-\frac{1}{2})^2 + y^2 + 2(-\frac{1}{2}) + 3y + 1 = y^2 + 3y + \frac{1}{4} = 0$. The equation of the circle with diameter $AB$ is $S_1 + k(S_1 - S_2) = 0$. Substituting $x = -\frac{1}{2}$ into $S_1$ and $S_2$ gives the points $A$ and $B$. The circle with diameter $AB$ is $x^2+y^2+2x+3y+1 + k(2x+1) = 0$. Since the center lies on the line $x = -\frac{1}{2}$,we find $k$ such that the equation simplifies to $x^2+y^2+x+3y+\frac{1}{2} = 0$,which is $2x^2+2y^2+2x+6y+1 = 0$.

If the circle $x^2+y^2+2x+3y+1=0$ cuts another circle $x^2+y^2+4x+3y+2=0$ at points $A$ and $B$,then the equation of the circle with $AB$ as a diameter is

Similar Questions

The length of the common chord of the circles $x^2 + y^2 + 2x + 3y + 1 = 0$ and $x^2 + y^2 + 4x + 3y + 2 = 0$ is

The length of the common chord of the two circles $x^2+y^2-4y=0$ and $x^2+y^2-8x-4y+11=0$ is

$L_1$ and $L_2$ are two common tangents to two circles. If $L_1$ touches the two circles at $A(1, 1)$ and $B(0, 1)$ and $L_2$ touches the two circles at $C\left(\frac{3}{5}, \frac{4}{5}\right)$ and $D\left(-\frac{1}{5}, \frac{7}{5}\right)$,then the equation of the radical axis of the two circles is

The area of the triangle formed by the tangents from the point $(h, k)$ to the circle $x^2 + y^2 = a^2$ and the line joining their points of contact is

If the chord of contact of tangents from a point $A$ to a given circle passes through $B$,then the circle with $AB$ as a diameter will . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module