The centre of a circle which cuts x^{2}+y^{2} | vedclass

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

Question

(D) Let the required circle be $S: x^{2}+y^{2}+2gx+2fy+c=0$. For a circle $x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$,the condition for orthogonality is $2g

Vedclass · Accepted Answer

$\left(-\frac{1}{7}, \frac{9}{7}\right)$

Vedclass · Answer

(D) Let the required circle be $S: x^{2}+y^{2}+2gx+2fy+c=0$. For a circle $x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$,the condition for orthogonality is $2gg_{i}+2ff_{i}=c+c_{i}$. For the given circles: $S_{1}: x^{2}+y^{2}+6x+0y-1=0 \implies 2g(3)+2f(0)=c-1 \implies 6g=c-1$ $(i)$ $S_{2}: x^{2}+y^{2}+0x-3y+2=0 \implies 2g(0)+2f(-3/2)=c+2 \implies -3f=c+2$ (ii) $S_{3}: x^{2}+y^{2}+x+y-3=0 \implies 2g(1/2)+2f(1/2)=c-3 \implies g+f=c-3$ (iii) Subtracting (ii) from $(i)$: $6g+3f=-3 \implies 2g+f=-1$ (iv) Subtracting (iii) from $(i)$: $5g-f=2$ $(v)$ Adding (iv) and $(v)$: $7g=1 \implies g=1/7$. Substituting $g$ in (iv): $2(1/7)+f=-1 \implies f=-1-2/7 = -9/7$. The centre of the circle is $(-g, -f) = (-1/7, 9/7)$.

The centre of a circle which cuts $x^{2}+y^{2}+6x-1=0$,$x^{2}+y^{2}-3y+2=0$ and $x^{2}+y^{2}+x+y-3=0$ orthogonally is

Similar Questions

If $A$ and $B$ are the centres of similitude with respect to the circles $x^2+y^2-14x+6y+33=0$ and $x^2+y^2+30x-2y+1=0$,then the midpoint of $AB$ is

If the circles $x^2+y^2+kx+4y+2=0$ and $2(x^2+y^2)-4x-3y+k=0$ cut orthogonally,then $k=$

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

The number of common tangents to the two circles $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$ is:

Find the equation of a circle with radius $5$ units and touching the circle $x^2+y^2-2x-4y-20=0$ at the point $(5,5)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module