If the radical axis of the circles x^2+y^2+2g | vedclass

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

Question

(A) The equation of the first circle is $x^2+y^2+2gx+2fy+c=0$. The equation of the second circle is $2x^2+2y^2+3x+8y+2c=0$,which can be written as $x^

Vedclass · Accepted Answer

either $g=\frac{3}{4}$ or $f=2$

Vedclass · Answer

(A) The equation of the first circle is $x^2+y^2+2gx+2fy+c=0$. The equation of the second circle is $2x^2+2y^2+3x+8y+2c=0$,which can be written as $x^2+y^2+\frac{3}{2}x+4y+c=0$. The radical axis is obtained by subtracting the two equations: $(2g-\frac{3}{2})x + (2f-4)y = 0$. This line passes through the origin $(0,0)$. The third circle is $x^2+y^2+2x+2y+1=0$,which is $(x+1)^2+(y+1)^2=1$. Its center is $(-1,-1)$ and radius is $1$. The line $(2g-\frac{3}{2})x + (2f-4)y = 0$ touches this circle if the perpendicular distance from $(-1,-1)$ to the line is equal to the radius $1$. $\frac{|-(2g-\frac{3}{2})-(2f-4)|}{\sqrt{(2g-\frac{3}{2})^2+(2f-4)^2}} = 1$. Squaring both sides: $(-2g+\frac{3}{2}-2f+4)^2 = (2g-\frac{3}{2})^2+(2f-4)^2$. Let $A = 2g-\frac{3}{2}$ and $B = 2f-4$. Then $(-A-B)^2 = A^2+B^2$,which implies $A^2+B^2+2AB = A^2+B^2$,so $2AB=0$. Thus,$A=0$ or $B=0$. $2g-\frac{3}{2}=0 \implies g=\frac{3}{4}$ or $2f-4=0 \implies f=2$.

If the radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$,then

Similar Questions

The equation of the line joining the centres of the circles belonging to the coaxial system of circles $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$ is

The circles $x^2 + y^2 - 2x - 4y = 0$ and $x^2 + y^2 - 8y - 4 = 0$:

If $(-1, -1)$ is the radical centre of the circles $x^2 + y^2 + 2gx - 4y + 4 = 0$,$x^2 + y^2 + 6x + 2fy + 12 = 0$,and $x^2 + y^2 + 10y + 20 = 0$,then $g - f = $

If the circles $x^{2}+y^{2}+2x+2ky+6=0$ and $x^{2}+y^{2}+2ky+k=0$ intersect orthogonally,then $k$ is

The point at which the circles $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$ touch each other is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module