The equation of the circle,the end-points of | vedclass

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

Question

(A) The center of the first circle $x^{2}+y^{2}-2x+3y-3=0$ is found by comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,giving $C_{1} = (-g, -f) = (1, -3/2)$.

Vedclass · Accepted Answer

$2x^{2}+2y^{2}+4x-9y-24=0$

Vedclass · Answer

(A) The center of the first circle $x^{2}+y^{2}-2x+3y-3=0$ is found by comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,giving $C_{1} = (-g, -f) = (1, -3/2)$.The center of the second circle $x^{2}+y^{2}+6x-12y-5=0$ is $C_{2} = (-3, 6)$.The equation of a circle with diameter endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) = 0$.Substituting the centers $(1, -3/2)$ and $(-3, 6)$:$(x-1)(x+3) + (y+3/2)(y-6) = 0$$x^{2}+2x-3 + y^{2}-6y+3/2y-9 = 0$$x^{2}+y^{2}+2x-9/2y-12 = 0$Multiplying by $2$ to match the options:$2x^{2}+2y^{2}+4x-9y-24 = 0$.

The equation of the circle,the end-points of whose diameter are the centres of the circles $x^{2}+y^{2}-2x+3y-3=0$ and $x^{2}+y^{2}+6x-12y-5=0$ is

Similar Questions

The equation of a circle whose centre is the origin and whose radius is equal to the distance between the lines $x = 1$ and $x = -1$ is:

If $(3,1)$ and $(-2,4)$ are points on a circle $S$ whose centre lies on the line $x-y+1=0$,then the parametric equations of $S$ are:

The equations of the two circles which touch the $Y$-axis at $(0,3)$ and make an intercept of $8$ units on the $X$-axis are

If the incenter of an equilateral triangle is $(1, 1)$ and the equation of one of its sides is $3x + 4y + 3 = 0$,then the equation of the circumcircle of this triangle is

The equation of the circle that touches the $Y-$axis at a distance of $4$ units from the origin and cuts off an intercept of $6$ units on the $X-$axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module