The equation of the line joining the centres | vedclass

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

Question

(A) The given coaxial system of circles is $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$.Dividing by $4$,we get $x^2 + y^2 - 3x + \frac{3}{2}

Vedclass · Accepted Answer

$8x - 4y - 15 = 0$

Vedclass · Answer

(A) The given coaxial system of circles is $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$.Dividing by $4$,we get $x^2 + y^2 - 3x + \frac{3}{2}y - \frac{3}{4} + \frac{\lambda}{4}(x + 2y - 6) = 0$.The radical axis is $x + 2y - 6 = 0$. The line of centres is perpendicular to the radical axis,so it is of the form $2x - y + k = 0$.The centre of the base circle $x^2 + y^2 - 3x + \frac{3}{2}y - \frac{3}{4} = 0$ is $(\frac{3}{2}, -\frac{3}{4})$.Since the line of centres passes through this point,we substitute it into $2x - y + k = 0$:$2(\frac{3}{2}) - (-\frac{3}{4}) + k = 0 \implies 3 + \frac{3}{4} + k = 0 \implies k = -\frac{15}{4}$.Thus,the equation of the line of centres is $2x - y - \frac{15}{4} = 0$,which simplifies to $8x - 4y - 15 = 0$.

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

Similar Questions

If two circles which pass through the points $(0, a)$ and $(0, -a)$ and touch the line $y = mx + c$ cut orthogonally, then:

If the angle between the circles $x^2+y^2-4x-6y+k=0$ and $x^2+y^2+8x-4y+11=0$ is $\frac{\pi}{3}$,then a value of $k$ is

The equation of the radical axis of the pair of circles $7x^2+7y^2-7x+14y+18=0$ and $4x^2+4y^2-7x+8y+20=0$ is

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

The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-6x+8=0$ and $x^2+y^2-2x-2y-7=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module