If a circle passes through the point $(a , b) \&$ cuts the circle $x^2 + y^2= K^2$ orthogonally, then the equation of the locus of its centre is :
$2ax + 2by - (a^2 + b^2 + K^2) = 0$
$2ax + 2by - (a^2 - b^2+ K^2) = 0$
$x^2 + y^2 - 3ax - 4by + (a^2 + b^2 - K^2) = 0$
$x^2 + y^2 - 2ax - 3by + (a^2 - b^2 - K^2) = 0$
The distance from the centre of the circle $x^2 + y^2 = 2x$ to the straight line passing through the points of intersection of the two circles $x^2 + y^2 + 5x -8y + 1 =0$ and $x^2 + y^2-3x + 7y -25 = 0$ is-
Radical axis of the circles $3{x^2} + 3{y^2} - 7x + 8y + 11 = 0$ and ${x^2} + {y^2} - 3x - 4y + 5 = 0$ is
Let $Z$ be the set of all integers,
$\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}:(\mathrm{x}-2)^{2}+\mathrm{y}^{2} \leq 4\right\}$
$\mathrm{B}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}: \mathrm{x}^{2}+\mathrm{y}^{2} \leq 4\right\} \text { and }$
$\mathrm{C}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}:(\mathrm{x}-2)^{2}+(\mathrm{y}-2)^{2} \leq 4\right\}$
If the total number of relation from $\mathrm{A} \cap \mathrm{B}$ to $\mathrm{A} \cap \mathrm{C}$ is $2^{\mathrm{p}}$, then the value of $\mathrm{p}$ is :
Two circles of radii $4$ cms $\&\,\, 1\,\, cm$ touch each other externally and $\theta$ is the angle contained by their direct common tangents. Then $sin \theta =$
Two orthogonal circles are such that area of one is twice the area of other. If radius of smaller circle is $r$, then distance between their centers will be -