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

If $$\mathrm{P}(6,1)$$ be the orthocentre of the triangle whose vertices are $$\mathrm{A}(5,-2), \mathrm{B}(8,3)$$ and $$\mathrm{C}(\mathrm{h}, \mathrm{k})$$, then the point $$\mathrm{C}$$ lies on the circle :

A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are $$m$$ and $$n$$, respectively, then $$m+n^2$$ is equal to

Let the circle $$C_1: x^2+y^2-2(x+y)+1=0$$ and $$\mathrm{C_2}$$ be a circle having centre at $$(-1,0)$$ and radius 2 . If the line of the common chord of $$\mathrm{C}_1$$ and $$\mathrm{C}_2$$ intersects the $$\mathrm{y}$$-axis at the point $$\mathrm{P}$$, then the square of the distance of P from the centre of $$\mathrm{C_1}$$ is: