The equation of the circle which passes through the origin, has its centre on the line $x + y = 4$ and cuts the circle ${x^2} + {y^2} – 4x + 2y + 4 = 0$ orthogonally, is

Two circles whose radii are equal to $4$ and $8$ intersects at right angles. The length of  their common chord is:-

If the variable line $3 x+4 y=\alpha$ lies between the two circles $(x-1)^{2}+(y-1)^{2}=1$ and $(x-9)^{2}+(y-1)^{2}=4$ without intercepting a chord on either circle, then the sum of all the integral values of $\alpha$ is …. .

Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2: 1$, then $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$ equals

$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the co-ordinate axes cut at right angles, then :