If the circle C_1 : x^{2} + y^{2} = 16 inters | vedclass

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

Question

(C) Let the center of circle $C_2$ be $(h, k)$. Its equation is $(x - h)^{2} + (y - k)^{2} = 5^{2} = 25$.The equation of $C_1$ is $x^{2} + y^{2} = 16$

Vedclass · Accepted Answer

$(\pm 9/5, \pm 12/5)$

Vedclass · Answer

(C) Let the center of circle $C_2$ be $(h, k)$. Its equation is $(x - h)^{2} + (y - k)^{2} = 5^{2} = 25$.The equation of $C_1$ is $x^{2} + y^{2} = 16$.The equation of the common chord is given by $S_1 - S_2 = 0$,which is $2hx + 2ky = h^{2} + k^{2} - 9$.The length of the common chord is maximum when the chord passes through the center of $C_1$,i.e.,$(0, 0)$.Thus,$h^{2} + k^{2} - 9 = 0$,or $h^{2} + k^{2} = 9$.The slope of the common chord $2hx + 2ky = 9$ is $-h/k = 3/4$,which implies $k = -4h/3$.Substituting $k$ into $h^{2} + k^{2} = 9$,we get $h^{2} + (-4h/3)^{2} = 9$,so $h^{2} + 16h^{2}/9 = 9$,which gives $25h^{2}/9 = 9$,so $h^{2} = 81/25$.Thus,$h = \pm 9/5$. If $h = 9/5$,then $k = -4(9/5)/3 = -12/5$. If $h = -9/5$,then $k = 12/5$.Therefore,the possible centers are $(9/5, -12/5)$ and $(-9/5, 12/5)$.

If the circle $C_1 : x^{2} + y^{2} = 16$ intersects another circle $C_2$ of radius $5$ such that the length of the common chord is maximum and its slope is $3/4$,then the coordinates of the center of $C_2$ are:

Similar Questions

Tangents $OP$ and $OQ$ are drawn from the origin $O$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$. Then,the equation of the circumcircle of the triangle $OPQ$ is

Find the distance from the center of the circle $x^2 + y^2 = 2x$ to the common chord of the circles $x^2 + y^2 + 5x - 8y + 1 = 0$ and $x^2 + y^2 - 3x + 7y - 25 = 0$.

If the circles $x^{2}+y^{2}-2x-2y-7=0$ and $x^{2}+y^{2}+4x+2y+k=0$ cut orthogonally,then the length of the common chord of the circles is

The equation of the pair of tangents drawn from the origin to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is:

If the circle $S \equiv x^2+y^2-4=0$ intersects another circle $S^{\prime}=0$ of radius $\frac{5 \sqrt{2}}{2}$ in such a manner that the common chord is of maximum length with slope equal to $\frac{1}{4}$,then the centre of $S^{\prime}=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module