If one of the diameters of the circle given b | vedclass

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

Question

(A) The given circle is $x^{2}+y^{2}+4x+6y-12=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=2, f=3, c=-12$. The centre $C_{1}$ is $(-g, -f) =

Vedclass · Accepted Answer

$\sqrt{41} 	ext{ unit}$

Vedclass · Answer

(A) The given circle is $x^{2}+y^{2}+4x+6y-12=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=2, f=3, c=-12$. The centre $C_{1}$ is $(-g, -f) = (-2, -3)$ and the radius $r_{1} = \sqrt{g^{2}+f^{2}-c} = \sqrt{2^{2}+3^{2}-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5 	ext{ units}$. Since a diameter of this circle is a chord of circle $S$,the distance $d$ from the centre of $S$,$C_{2}(2, -3)$,to this chord is $0$ (because the chord passes through the centre of the first circle,but the problem states the diameter is a chord of $S$,implying the chord is a diameter of the first circle). Actually,the chord is a diameter of the first circle,so its length is $2r_{1} = 10$. The distance $d$ from $C_{2}(2, -3)$ to the centre $C_{1}(-2, -3)$ is $\sqrt{(2 - (-2))^{2} + (-3 - (-3))^{2}} = \sqrt{4^{2} + 0^{2}} = 4$. Let $R$ be the radius of circle $S$. The distance from the centre $C_{2}$ to the chord is $d=4$. The half-length of the chord is $a = r_{1} = 5$. Using the Pythagorean theorem in the triangle formed by the radius $R$,distance $d$,and half-chord $a$: $R^{2} = d^{2} + a^{2} = 4^{2} + 5^{2} = 16 + 25 = 41$. Therefore,$R = \sqrt{41} 	ext{ units}$.

If one of the diameters of the circle given by the equation $x^{2}+y^{2}+4x+6y-12=0$ is a chord of a circle $S$ whose centre is $(2,-3)$,then the radius of $S$ is:

Similar Questions

Consider the circles ${x^2} + {(y - 1)^2} = 9$ and ${(x - 1)^2} + {y^2} = 25$. They are such that:

$x^2+y^2+2x-6y-6=0$ and $x^2+y^2-6x-2y+k=0$ are two intersecting circles and $k$ is not an integer. If $\theta$ is the angle between the two circles and $\cos \theta = \frac{-5}{24}$,then $k=$

If one end of a diameter of the circle $x^2 + y^2 + 2x + 4y - 3 = 0$ is $(1, 0)$,then the other end of the diameter is:

If $A(-2, 1)$,$B(0, -2)$,and $C(1, 2)$ are the vertices of a triangle $ABC$,then the perpendicular distance from its circumcentre to the side $BC$ is

If $\theta$ is the angle between the tangents from $(-1, 0)$ to the circle $x^2+y^2-5x+4y-2=0$,then $\theta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module