The equation of the circle with center (-a, - | vedclass

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

Question

(C) The standard equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^{2} + (y-k)^{2} = r^{2}$.Given center $(h, k) = (-a, -b)$

Vedclass · Accepted Answer

$x^{2}+y^{2}+2ax+2by+2b^{2}=0$

Vedclass · Answer

(C) The standard equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^{2} + (y-k)^{2} = r^{2}$.Given center $(h, k) = (-a, -b)$ and radius $r = \sqrt{a^{2}-b^{2}}$.Substituting these values into the formula:$(x - (-a))^{2} + (y - (-b))^{2} = (\sqrt{a^{2}-b^{2}})^{2}$$(x+a)^{2} + (y+b)^{2} = a^{2}-b^{2}$Expanding the squares:$x^{2} + 2ax + a^{2} + y^{2} + 2by + b^{2} = a^{2} - b^{2}$Subtracting $a^{2}$ from both sides and adding $b^{2}$ to both sides:$x^{2} + y^{2} + 2ax + 2by + b^{2} + b^{2} = 0$$x^{2} + y^{2} + 2ax + 2by + 2b^{2} = 0$.

The equation of the circle with center $(-a, -b)$ and radius $\sqrt{a^{2}-b^{2}}$ is:

Similar Questions

The equation of the circle which touches both axes and whose centre is $({x_1}, {y_1})$ is

If the equation of the circle lying in the first quadrant,touching both the coordinate axes and the line $\frac{x}{3}+\frac{y}{4}=1$ is $(x-c)^2+(y-c)^2=c^2$,then $c=$

The radius of the circle $(x - 1)(x - 3) + (y - 2)(y - 4) = 0$ is

If the lines $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$ are two diameters of a circle of area $49\pi$ square units,the equation of the circle is:

The four distinct points $(0, 0), (2, 0), (0, -2)$ and $(k, -2)$ are concyclic,if $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module