Suppose the circle S: x^2+y^2+2gx+2fy+c=0 cut | vedclass

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

Question

(C) The condition for two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to cut orthogonally is $2g_1g_2+2f_1f_2=c_1+c_2$. For $S

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The condition for two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to cut orthogonally is $2g_1g_2+2f_1f_2=c_1+c_2$. For $S$ and $S'$,we have $2g(-2)+2f(-3)=c+11 \implies -4g-6f=c+11$ $(i)$. For $S$ and $S''$,we have $2g(-5)+2f(-2)=c+21 \implies -10g-4f=c+21$ $(ii)$. The centre of $S$ is $(-g, -f)$. Since it lies on the bisector of the angle between positive coordinate axes $(y=x)$,we have $-f = -g$,so $f=g$ $(iii)$. Substituting $f=g$ into $(i)$ and $(ii)$: $-10f = c+11$ $(iv)$ $-14f = c+21$ $(v)$ Subtracting $(v)$ from $(iv)$: $4f = -10 \implies f = -2.5$. Thus $g = -2.5$. From $(iv)$,$c = -10(-2.5) - 11 = 25 - 11 = 14$. Finally,$2g+2f+c = 2(-2.5)+2(-2.5)+14 = -5-5+14 = 4$.

Suppose the circle $S: x^2+y^2+2gx+2fy+c=0$ cuts orthogonally the two circles $S': x^2+y^2-4x-6y+11=0$ and $S'': x^2+y^2-10x-4y+21=0$. If the centre of $S=0$ lies on the bisector of the angle between the positive coordinate axes,then $2g+2f+c=$

Similar Questions

The equation of the circle passing through the points of intersection of the circles $x^2+y^2-2px=0$ and $x^2+y^2-2qy=0$ and having its centre on the line $\frac{x}{p}-\frac{y}{q}=2$ is:

If $y + 3x = 0$ is the equation of a chord of the circle $x^2 + y^2 - 30x = 0$,then the equation of the circle with this chord as diameter is:

The line $x+y+2=0$ intersects the circle $x^2+y^2+4x-4y-4=0$ at two points $A$ and $B$. Let $S \equiv x^2+y^2+2gx+2fy+c=0$ be a different circle passing through the points $A$ and $B$. If the distance of the centre of $S=0$ from $AB$ is $\sqrt{2}$,then $g+f+c=$

Where do the circles $x^2 + y^2 + 2x - 2y + 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ touch each other?

One of the limit points of the coaxial system of circles containing $x^2 + y^2 - 6x - 6y + 4 = 0$ and $x^2 + y^2 - 2x - 4y + 3 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module