The equation of a circle passing through (-6, | vedclass

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

Question

(C) Since the circle touches both coordinate axes and passes through the point $(-6, 3)$ in the second quadrant,its center must be $(-r, r)$ and its r

Vedclass · Accepted Answer

$x^2+y^2+6x-6y+9=0$

Vedclass · Answer

(C) Since the circle touches both coordinate axes and passes through the point $(-6, 3)$ in the second quadrant,its center must be $(-r, r)$ and its radius must be $r$,where $r > 0$.The equation of the circle is $(x+r)^2 + (y-r)^2 = r^2$.Expanding this,we get $x^2 + 2xr + r^2 + y^2 - 2yr + r^2 = r^2$,which simplifies to $x^2 + y^2 + 2xr - 2yr + r^2 = 0$.Since the circle passes through $(-6, 3)$,we substitute these coordinates into the equation:$(-6)^2 + (3)^2 + 2(-6)r - 2(3)r + r^2 = 0$$36 + 9 - 12r - 6r + r^2 = 0$$r^2 - 18r + 45 = 0$$(r - 3)(r - 15) = 0$Thus,$r = 3$ or $r = 15$.For $r = 3$,the equation is $x^2 + y^2 + 2(3)x - 2(3)y + (3)^2 = 0$,which is $x^2 + y^2 + 6x - 6y + 9 = 0$.For $r = 15$,the equation is $x^2 + y^2 + 30x - 30y + 225 = 0$.

The equation of a circle passing through $(-6, 3)$ and touching both the coordinate axes is

Similar Questions

Find the equation of the circle which passes through the points $(2, -2)$ and $(3, 4)$ and whose centre lies on the line $x + y = 2$.

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154$ square units. The equation of the circle is

The radius of the circle $r=12 \cos \theta+5 \sin \theta$ is

The coordinates of any point on the circle $x^2 + y^2 = 4$ are:

The equation of the circle concentric with the circle $x^2 + y^2 - 6x + 7 = 0$ and which touches the line $x + y + 3 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module