Family of Circles

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

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

Consider a family of circles passing through two fixed points $A\left(3,7\right)$ and $B\left(6,5\right)$. If the common chords of the circle $x^2+y^2-4x-6y-3=0$ and the members of the family of circles pass through a fixed point $\left(a,b\right)$, then

$P$ is a point on the parabola $y^2=16x$ where abscissa and ordinate are equal. Equation of a circle passing through the focus and touching the parabola at $P$ is

Consider a family of circles passing through two fixed points $A\left(3,7\right)$ and $B\left(6,5\right)$. The chord in which the circle $x^2+y^2-4x-6y-3=0$ cuts each member of family of circles passes through a fixed point $\left(a,b\right)$. Then the value of $a+3b$ is

If $x-y+1=0$ meets the circle $x^2+y^2+y-1=0$ at A and B then the equation of the circle with AB as diameter is

Find the equation of the circle through the point $\left(-2, 4\right)$ and through the points of intersection of the circle $x^2+y^2+2x-4y-4=0$ and the line $3x+2y-5=0$

Equation of the circle, whose diameter is the chord $x+y=1$ of the circles $x^2+y^2=4$ , is -

Equation of smallest circle passing through 'points of intersection of line $x+y=1$ & circle $x^2+y^2=9$

The equation of the circle whose diameter is the common chord of the circles ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}+3\mathrm{y}+2=0$ and ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-3\mathrm{y}-\mathrm{ }4=0$ is-

The circle whose diameter is the chord $\mathrm{x}+\mathrm{y}=1$ of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=4$ is

The intercept on the line $y=x$ by the circle $x^2+y^2-2x=0$ is $AB\mathit{.}$ Equation of circle with $AB$ as diameter is-

The equation of circle passing through the points of intersection of circles $x^2+y^2-6x+8=0$ and $x^2+y^2=6$ and point $\left(\mathrm{1,1}\right)$ is -

Equation of the circle, whose diameter is the chord $x+y=1$ of the circles $x^2+y^2=4$ , is -

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 the conics whose equations are

${\mathrm{S}}_1\mathrm{ }:\mathrm{ }\left({\sin }^2\mathrm{\theta }\right){\mathrm{x}}^2+\left(2\mathrm{h}\tan \mathrm{\theta }\right)\mathrm{x}\mathrm{y}+\left({\cos }^2\mathrm{\theta }\right){\mathrm{y}}^2+32\mathrm{x}+16\mathrm{y}+19=0$

${\mathrm{S}}_2\mathrm{ }:\mathrm{ }\left({\cos }^2\mathrm{\theta }\right){\mathrm{x}}^2-\left(2\mathrm{h}\mathrm{\prime }\cot \mathrm{\theta }\right)\mathrm{x}\mathrm{y}+\left({\sin }^2\mathrm{\theta }\right){\mathrm{y}}^2+16\mathrm{x}+32\mathrm{y}+19=0$

intersect in four concyclic points, where $\mathrm{\theta }\in \mathrm{ }\left[0,\mathrm{ }\frac{\mathrm{\pi }}{2}\right]$ , then the correct statement(s) can be

Consider the equation of the circle described on the common chord of circles $x^2+y^2-4x-5=0$ and $x^2+y^2+8y+7=0$ as a diameter, then

Consider the equation of the circle described on the common chord of circles x² + y² – 4x – 5 = 0 and x² + y² + 8y + 7 = 0 as a diameter, then

Equation of circle through intersection of $x^2+y^2+2x= 0$ and $x-y=0$, having minimum radius is -

The equation of the circle which passes through the intersection of $x^2+y^2+13x-3y=0$ and $2x^2+2y^2+4x-7y-25=0$ and whose centre lies on $13x+30y=0$, is