The equation of the circle whose diameter is | vedclass

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

Question

(A) Given circles are $S_1 \equiv x^2+y^2+2x+2y+1=0$ and $S_2 \equiv x^2+y^2+4x+6y+4=0$. The equation of the common chord is $S_1 - S_2 = 0$: $(x^2+y^

Vedclass · Accepted Answer

$10x^2+10y^2+14x+8y+1=0$

Vedclass · Answer

(A) Given circles are $S_1 \equiv x^2+y^2+2x+2y+1=0$ and $S_2 \equiv x^2+y^2+4x+6y+4=0$. The equation of the common chord is $S_1 - S_2 = 0$: $(x^2+y^2+2x+2y+1) - (x^2+y^2+4x+6y+4) = 0$ $-2x - 4y - 3 = 0 \Rightarrow 2x + 4y + 3 = 0$. The equation of a circle passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$ (or $S_1 + \lambda(L) = 0$ where $L$ is the common chord). $x^2+y^2+2x+2y+1 + \lambda(2x+4y+3) = 0$ $x^2+y^2+2x(1+\lambda) + 2y(1+2\lambda) + (1+3\lambda) = 0$. The center of this circle is $(- (1+\lambda), -(1+2\lambda))$. Since the common chord $2x+4y+3=0$ is a diameter,the center must lie on it: $2(-(1+\lambda)) + 4(-(1+2\lambda)) + 3 = 0$ $-2 - 2\lambda - 4 - 8\lambda + 3 = 0$ $-10\lambda - 3 = 0 \Rightarrow \lambda = -\frac{3}{10}$. Substituting $\lambda = -\frac{3}{10}$ into the circle equation: $x^2+y^2+2x(1-\frac{3}{10}) + 2y(1-\frac{6}{10}) + (1-\frac{9}{10}) = 0$ $x^2+y^2+\frac{14x}{10} + \frac{8y}{10} + \frac{1}{10} = 0$ $10x^2+10y^2+14x+8y+1=0$.

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2+4x+6y+4=0$ is

Similar Questions

Find the distance between the chords of contact of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ with respect to the points $(0, 0)$ and $(g, f)$.

If the chord of contact of tangents from a point on the circle $x^2+y^2=r_1^2$ to the circle $x^2+y^2=r_2^2$ touches the circle $x^2+y^2=r_3^2$,then $r_1, r_2, r_3$ are in:

Find the condition that the pair of lines joining the origin to the points of intersection of the line $y = mx + c$ and the circle $x^{2} + y^{2} = a^{2}$ are at right angles to each other.

If tangents are drawn to the circle $x^2+y^2=12$ at the points of intersection with the circle $x^2+y^2-5x+3y-2=0$,then the ordinate of the point of intersection of these tangents is

The equation of the pair of tangents drawn from the origin to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module