The equation of the circle which passes throu | vedclass

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

Question

(B) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$ $(i)$.Since the circle passes through $(3,2)$,we have $9+4+6g+4f+c=0$,which impli

Vedclass · Accepted Answer

$x^2+y^2+6x-8y-15=0$

Vedclass · Answer

(B) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$ $(i)$.Since the circle passes through $(3,2)$,we have $9+4+6g+4f+c=0$,which implies $6g+4f+c+13=0$ $(ii)$.Since the circle bisects the circumference of $x^2+y^2=15$,the common chord is the diameter of $x^2+y^2=15$. The radical axis is $2gx+2fy+c-15=0$. Since the center $(0,0)$ lies on this line,we get $c-15=0$,so $c=15$ is incorrect; rather,the condition for bisecting the circumference is that the radical axis passes through the center $(0,0)$,so $c-15=0$ is wrong. The radical axis is $2gx+2fy+c+15=0$. Thus $c+15=0$,so $c=-15$ $(iii)$.Since the circle cuts $x^2+y^2+4x+6y+3=0$ orthogonally,we use $2g_1g_2+2f_1f_2=c_1+c_2$. Here $g_1=g, f_1=f, c_1=c$ and $g_2=2, f_2=3, c_2=3$. So $2g(2)+2f(3)=c+3$,which gives $4g+6f=c+3$ $(iv)$.Substituting $c=-15$ in $(iv)$,$4g+6f=-12$,so $2g+3f=-6$.Substituting $c=-15$ in $(ii)$,$6g+4f=2$,so $3g+2f=1$.Solving these,$g=3, f=-4$. Thus the equation is $x^2+y^2+6x-8y-15=0$.

The equation of the circle which passes through the point $(3,2)$,bisects the circumference of the circle $x^2+y^2=15$,and cuts the circle $x^2+y^2+4x+6y+3=0$ orthogonally is

Similar Questions

If the circle $x^2+y^2+6x-2y+k=0$ bisects the circumference of the circle $x^2+y^2+2x-6y-15=0$,then $k$ is equal to :

If the lengths of the tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the coordinates of point $P$ are:

Any circle passes through the point of intersection of the lines $x + \sqrt{3}y = 1$ and $\sqrt{3}x - y = 2$. If it intersects these lines at points $P$ and $Q$,then the angle subtended by the arc $PQ$ at its centre is ............ $^o$.

$A$ circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then
$(A)$ radius of $S$ is $8$
$(B)$ radius of $S$ is $7$
$(C)$ centre of $S$ is $(-7,1)$
$(D)$ centre of $S$ is $(-8,1)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module