Angle between Two Circles

The equation of the common chord of the pair of circles: $(x-a{)}^2+(y-b{)}^2=c^2,$ and $(x-b{)}^2+(y-a{)}^2=c^2(a\ne b)$ is

The equation of the circle which cuts orthogonally the circle $x^2+y^2-4x+2y-7=0$ and having the centre at $(2,3)$ is

The equation of the circle which passes through the points $(2,0),(0,2)$ and orthogonal to the circle $2x^2+2y^2+5x-6y+4=0$ is

The equation of the circle passing through the origin, having its centre on the line $x+y=4$ and intersecting the circle $x^2+y^2-4x+2y+4=0$ orthogonally is

The equation of the circle which passes through the origin and intersects the circles, $x^2+y^2-4x-6y-3=0,$ and $x^2+y^2-8y+12=0$ orthogonally is

The equation of the circle which passes through the origin and intersects the circles, orthogonally: $x^2+y^2-4x+6y+10=0,$ and $x^2+y^2+12y+6=0$ is

The angle between the circles given by the equations: $x^2+y^2+6x-10y-135=0, \text{and} x^2+y^2-4x+14y-116=0$ is

If the angle between the circles given by the equations: $x^2+y^2-12x-6y+41=0,x^2+y^2+4x+6y-59=0$ is $\frac{\mathrm{\pi }}{k}$, then write the value of $k$.

Find $\text{'}k\text{'}$ if the following pairs of circles are orthogonal: $x^2+y^2+4x+8=0,x^2+y^2-16y+k=0$.

Find $\text{'}k\text{'}$ if the following pairs of circles are orthogonal: $x^2+y^2-5x-14y-34=0,x^2+y^2+2x+4y+k=0$.

Find $\text{'}k\text{'}$ if the following pairs of circles are orthogonal: $x^2+y^2-6x-8y+12=0, x^2+y^2-4x+6y+k=0$.

Find $\text{'}k\text{'}$ if the following pairs of circles are orthogonal: $x^2+y^2+2by-k=0,x^2+y^2+2ax+8=0$.

In List-$\mathrm{I}$, a pair of circles is given in $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and in List-$\mathrm{II}$, angle between those pair of circles is given. Match the items from List-$\mathrm{I}$ to List-$\mathrm{II}$

The possible number of values of $a$ for which the common chord of the circles $x^2+y^2=8$ and ${\left(x-a\right)}^2+y^2=8$ subtends a right angle at the origin is

Let $\mathrm{P}\mathrm{Q}$ be the common chord of the circles $S_1:x^2+y^2+2x+3y+1=0$ and $S_2:x^2+y^2+4x+3y+2=0$, then the perimeter (in units) of the triangle ${\mathrm{C}}_1\mathrm{P}\mathrm{Q}$ is equal to $\left(\mathrm{where},\hspace{0.17em}{C}_1=\left(-1,\frac{-3}{2}\right)\right)$

The locus of the centre of the circle which cuts the parabola $y^2=4x$ orthogonally at $(1, 2)$ will pass through the point

If the circles $x^2+y^2-2\lambda x-2y-7=0$ and $3\left(x^2+y^2\right)-8x+29y=0$ are orthogonal then $\lambda =$

If the angle between the circles $x^2+y^2-2x-4y+c=0$ and $x^2+y^2-4x-2y+4=0$ is $60°$, then $c$ is equal to

The length of the common chord of the two circles ${\left(x-a\right)}^2+y^2=a^2 and x^2+{\left(y-b\right)}^2=b^2$ is

The length of the common chord of the two circles ${\left(x-a\right)}^2+y^2=a^2 and x^2+{\left(y-b\right)}^2=b^2$ is