If one of the diameters of the circle x^2+y^2 | vedclass

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

Question

(C) The given circle is $x^2+y^2-10x+4y+13=0$. Its center $M$ is $(5, -2)$ and its radius $r$ is $\sqrt{5^2+(-2)^2-13} = \sqrt{25+4-13} = \sqrt{16} =

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The given circle is $x^2+y^2-10x+4y+13=0$. Its center $M$ is $(5, -2)$ and its radius $r$ is $\sqrt{5^2+(-2)^2-13} = \sqrt{25+4-13} = \sqrt{16} = 4$.The center of circle $C$ is the intersection of $2x+3y=12$ and $3x-2y=5$. Solving these,we get $x=3$ and $y=2$. So,the center of $C$ is $O(3, 2)$.The diameter of the first circle is a chord of circle $C$. Let $P$ be a point on the circumference of circle $C$ such that $MP$ is the radius of the first circle,i.e.,$MP = 4$. Since $M$ is the center of the first circle and $MP$ is a radius,$M$ lies on the chord.The distance $d$ between the centers $O(3, 2)$ and $M(5, -2)$ is $d = \sqrt{(5-3)^2 + (-2-2)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4+16} = \sqrt{20}$.In the right-angled triangle formed by the center of $C$,the center of the first circle,and a point on the chord,the radius $R$ of circle $C$ is given by $R^2 = d^2 + r^2 = 20 + 16 = 36$. Thus,$R = 6$.

If one of the diameters of the circle $x^2+y^2-10x+4y+13=0$ is a chord of another circle $C$,whose center is the point of intersection of the lines $2x+3y=12$ and $3x-2y=5$,then the radius of the circle $C$ is

Similar Questions

If two circles touch at $(0,0)$ externally,then their radical axis is

The equation of the circle touching the line $2x+3y+1=0$ at the point $(1,-1)$ and orthogonal to the circle which has the line segment having end points $(0,-1)$ and $(-2,3)$ as diameter,is

The circle passing through the intersection of the circles $x^{2}+y^{2}-6x=0$ and $x^{2}+y^{2}-4y=0$,having its centre on the line $2x-3y+12=0$,also passes through the point:

If the smallest circle through the points of intersection of $x^2+y^2=a^2$ and $x \cos \alpha+y \sin \alpha=p$,where $0 < p < a$,is $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$,then $\lambda=$

The equation of the circle passing through the points of intersection of the two orthogonal circles $S_1 = x^2 + y^2 + kx - 4y - 1 = 0$ and $S_2 = 3x^2 + 3y^2 - 14x + 23y - 15 = 0$ and passing through the point $(-1, -1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module