Parabola to Circle

Since the coefficient of ${x}^{2}$ in the quadratic is positive, the parabola intersects the y-axis at a point of negative ordinate. Moreover, the two distinct points of intersection of the parabola with the x-axis have opposite signs. Thus, every circle ${{C}_{p,q}}$ has a second common point with the y-axis, which belongs to the positive half-axis $Oy$ . Label the four points where an arbitrary circle ${{C}_{p,q}}$ cuts the axes as shown in the figure. If ${{x}_{1}}$ , ${{x}_{2}}$ are the roots of the quadratic equation ${{x}^{2}}+2px+q=0$ , with ${{x}_{1}}<{{x}_{2}}$ , then the points are $A\left( 0,\ q \right), \ \ \ \ \ B\left( {{x}_{1}},\ 0 \right) , \ \ \ \ \ C\left( {{x}_{2}},\ 0 \right) \ \ \ and \ \ \ D\left( 0,\ {{y}_{D}} \right)$ Using the power of a point theorem (also called Intersecting Chords Theorem ) we have $OB\cdot OC=OA\cdot OD\Rightarrow \left( -{{x}_{1}} \right)\cdot {{x}_{2}}=-q\cdot {{y}_{D}}\Rightarrow {{y}_{D}}=\frac{q}{{{x}_{1}}{{x}_{2}}}\Rightarrow {{y}_{D}}=1 \ \ \ \ \ {\color{#20A900}{\text{by Vieta's Formula for Quadratics}}}$ This means that the fixed point $D\left( 0,\ 1 \right)$ belongs to any circle ${{C}_{p,q}}$ . For the answer, $a+b=0+1=\boxed{1}$ .

Starting with,

$y = x^2 + 2 p x + q = (x + p)^2 + q - p^2$

The x-intercepts are,

$P_1 = ( - p - \sqrt{p^2 - q} , 0 )$

$P_2 = ( - p + \sqrt{p^2 - q} , 0 )$

and the y-intercept is $P_3 = (0, q)$

The center of circle passing through $P_1$ , $P_2$ and $P_3$ is $C=(-p, h)$ , where $h$ is found by setting,

$|CP_1|^2 = |CP_3|^2$

that is,

$p^2 - q + h^2 = p^2 + h^2 - 2 h q + q^2$

Simplifying

$h = \frac{1}{2} (q + 1)$

So, now we have the center coordinates,

$C = (-p, \frac{1}{2} (q + 1) )$

Equation of the circle, is given by,

$(x + p)^2 + (y - \frac{1}{2} (q + 1) )^2 = r^2$

Plugging in $P_3 = (0, q)$ , we get

$p^2 + (\frac{1}{2} (q - 1) )^2 = r^2$

and hence, the equation of the circle for any p, and q is

$(x + p)^2 + (y - \frac{1}{2} (q + 1) )^2 = p^2 + (\frac{1}{2}(q - 1))^2$

The invariant point $(a, b)$ can be found by differentiating the circle equation with respect to p and q and setting the derivatives to zero.

Differentiating with respect to $p$

$2 (a + p) (\dfrac{\partial a}{\partial p} + 1) + 2 (b - \frac{1}{2} (q + 1) )(\dfrac{\partial b}{\partial p}) = 2 p$

Plugging in $\dfrac{\partial a}{\partial p} = \dfrac{\partial b}{\partial p} = 0$ , the above equation becomes,

$2 (a + p) = 2 p$

from which it follows that $a = 0$

Secondly, differentiating the circle equation with respect to $q$ ,

$2 (a + p) (\dfrac{\partial a}{\partial q}) + 2 ( b - \frac{1}{2} (q + 1) ) (\dfrac{\partial b}{\partial q} - \frac{1}{2} ) = \frac{1}{2}(q - 1)$

Plugging in $\dfrac{\partial a}{\partial q} = \dfrac{\partial b}{\partial q} = 0$ , results in,

$- (b - \frac{1}{2} (q + 1) ) = \frac{1}{2} (q - 1)$

from which, $b = 1$

Therefore, the common point is $(a, b) = (0, 1)$ , and the answer is $0 + 1 = \boxed{1}$ .

From $y=x^2+2px+q$ , the $y$ -intercept is when $x=0$ or $y_0 = q$ . And the $x$ -intercepts are given by $x^2 + 2px + q = 0$ or $x_0 = - p \pm \sqrt{p^2-q}$ . Let the center of circle $C_{pq}$ be $C(u,v)$ and its radius be $r$ . Then we have:

$\begin{cases} (u-0)^2 + (v-y_0)^2 = u^2 + (v-q)^2 = r^2 & ...(1) \\ (u-x_0)^2 + (v-0)^2 = \left(u+p \pm \sqrt{p^2-q}\right)^2 + v^2 = r^2 & ...(2) \end{cases}$

It is obvious that $u = \dfrac {x_0+x_1}2 = - p$ . And

$\begin{cases} (1): & p^2 + v^2 - 2qv + q^2 = r^2 & ...(3) \\ (2): & p^2 - q + v^2 = r^2 & ...(4) \end{cases}$

From $(3)-(4): \ q - 2qv + q^2 = 0 \implies v = \dfrac {1+q}2$ ; from $(1): \ r^2 = p^2 + \left(\dfrac {1-q}2\right)^2$ ; and the equation of $C_{pq}$ is:

$\begin{aligned} (x+p)^2 + \left(y - \frac {1+q}2 \right)^2 & = p^2 + \left(\frac {1-q}2 \right)^2 & \small \blue{\text{when }x=0} \\ p^2 + \left(y - \frac {1+q}2 \right)^2 & = p^2 + \left(\frac {1-q}2 \right)^2 \\ \left(y - \frac {1+q}2 \right)^2 & = \left(\frac {1-q}2 \right)^2 \\ y^2 - (1+q)y + \frac {1+2q+q^2}4 & = \frac {1-2q+q^2}4 \\ y^2 -(1+q)y + q & = 0 \\ (y-1)(y-q) & = 0 \end{aligned}$

So we note that the point $(0,1)$ is in every $C_{pq}$ . Therefore $a+b = 0+1 = \boxed 1$ .