If the straight line $ax + by = 2$ where $a, b \neq 0$ touches the circle $x^2 + y^2 - 2x = 3$ and is normal to the circle $x^2 + y^2 - 4y = 6$,then the values of $a$ and $b$ are respectively:

The locus of the centers of the circles such that the point $(2, 3)$ is the midpoint of the chord $5x + 2y = 16$ is:

Let $a$ and $b$ be two non-zero real numbers. The equation $(ax^2 + by^2 + c)(x^2 - 5xy + 6y^2) = 0$ represents:

Let $C$ be the circle $x^2+(y-1)^2=2$. Let $E_1$ and $E_2$ be two ellipses whose centers lie at the origin and whose major axes lie on the $x$-axis and $y$-axis,respectively. Let the straight line $x+y=3$ touch the curves $C$,$E_1$,and $E_2$ at $P(x_1, y_1)$,$Q(x_2, y_2)$,and $R(x_3, y_3)$,respectively. Given that $P$ is the midpoint of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$,the value of $9(x_1y_1 + x_2y_2 + x_3y_3)$ is equal to . . . . . . .

If two circles of the same radius $a$ and centers at $(2, 3)$ and $(5, 6)$ are orthogonal,find the value of $a$.

If the shortest distance between the parabola $y^2=4x$ and the circle $x^2+y^2-4x-16y+64=0$ is $d$,then $d^2$ is equal to: