System of Circles

The locus of the centers of the circles that are passing through the intersection of the circles $x^2+y^2=1$ and $x^2+y^2-2x+y=0$ is

Find the equation of a circle which passes through the point $(1, 2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0.$

The circle described on the chord $x\mathrm{ }\mathrm{cos\alpha }+y\mathrm{ }\mathrm{sin\alpha }-p=0$ of the circle $x^2+y^2=a^2$ as diameter passes through the origin if

If $x+y=3$ is the equation of the chord $AB$ of the circle $x^2+y^2-2x+4y-8=0$, then the equation of the circle having $AB$ as diameter is

Center of the circle which passes through the point $\left(-2,2\right)$ and touches the circle $x^2+y^2-6x+8y=0$ at the point $\left(0,0\right)$ will be

The locus of the centres of the circles which cut the circles $x^2+y^2+4x-6y+9=0$ and $x^2+y^2-5x+4y-2=0$ orthogonally is -

The point $\left(\right[P+1], [P\left]\right)$, (where, $\left[.\right]$ denotes the greatest integer function) inside the region bounded by the circle $x^2+y^2-2x-15=0$ and $x^2+y^2-2x-7=0$, then

The radius of the circle touching the pair of lines $7x^2-18xy+7y^2=0$ and the circle $x^2+y^2-8x-8y=0$ and is contained in the given circle is

If one common tangent of the two circles ${\mathit{\text{x}}}^2+{\mathit{\text{y}}}^2=4$ and $x^2+{\left(y-3\right)}^2=\lambda , \lambda >0$ passes through the point $\left(\sqrt{3},1\right)$, then possible value of $4\lambda$ is

For different values of $\lambda ,$ the circle $x^2+y^2+(8+\lambda )x+(8+\lambda )y+16+12\lambda =0$ passes through two fixed points $A$ and $B$. Value of $\lambda$ for which tangents at $A$ and $B$ to the circle intersect at origin, is

The circle $x^2+y^2+kx+\left(k+1\right)y-\left(k+1\right)=0$ always passes through two fixed point for every real $k$. If the minimum value of the radius of the circle is $\frac{1}{\sqrt{P}}$, then the value of $P$ is

A circle $S$ touches the line $x+y=2$ at $\left(1,1\right)$ and cuts the circle $x^2+y^2+4x+5y-6=0$ at $P$ and $Q$ respectively. The $PQ$ always passes through the point 

Consider a family of circle passing through the point of intersection of lines $\sqrt{3}\left(y-1\right)=x-1$ and $y-1=\sqrt{3}\left(x-1\right)$ and having its centre on the acute angle bisector of the given lines. Then the common chords of each member of the family and the circle $x^2+y^2+4x-6y+5=0$ are concurrent at the point

Locus of a variable point $P\left(\mathit{x}\mathit{,}\mathit{ }\mathit{y}\right)$ is given by $x^3+y^3+3xy=1$ where $\left(x,y\right)\ne \left(-1, -1\right)$. The equation of circle touching the locus of $P$ at $\left(-1, 2\right)$ and passing through $\left(1, -2\right)$ is given by

A variable circle always touches the line $x+y-2=0$ at $(1,1)$ and cuts the circle $x^2+y^2+4x+5y-4=0$ at $\mathrm{A}$ and $\mathrm{B}$. If the line joining $\mathrm{AB}$ always passes through fixed point $(\alpha ,\beta ),$ then $\left|\frac{4\beta }{\alpha }\right|$ is equal to

Three circle touches one another externally. The radius of circles are three consecutive integers. The tangent at their point of contact meet at a point whose distance from a point of contact is $4.$ If the ratio of radius of largest to smallest circle is $\frac{k}{6}$, then find $k$

The radius of the circle touching the line $\mathrm{x}+\mathrm{y}=4$ at $\left(\mathrm{1,3}\right)$ and intersecting ${\mathrm{x}}^2+{\mathrm{y}}^2=4$ orthogonally is

If first circle $C$ is passing through the point $\left(4, 0\right)$ which touches the second circle $x^2+y^2+4x-6y=12$ externally at the point $\left(1, -1\right),$ then the radius of first circle $C$ is:

If the diameter of first circle is the common chord for other circles having equation $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$, then find the equation of first circle

Choose the correct relation between the two circles having equations as $x^2+y^2+6x+6y=0$ and $x^2+y^2-12x-12y=0$.