Question

# If the abscissa and ordinates of two points P and Q are the roots of the equations ${{x}^{2}}+2ax-{{b}^{2}}=0\text{ and }{{x}^{2}}+2px-{{q}^{2}}=0$ respectively, then equation of the circle with PQ as diameter is:

Hint : In this question, we first need to find the roots of the given quadratic equations which gives us the abscissa and ordinates of the required diameter. Then by substituting these values of abscissa and ordinates in the formula of the circle equation when the coordinates of the end points of the diameter are given.

$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$

QUADRATIC EQUATION: In algebra, a quadratic equation is any equation that can be rearranged in standard form as where x represents an unknown and a, b, c represent known numbers, where $a\ne 0$.

If the roots of the quadratic equation $a{{x}^{2}}+bx+c=0\left( a\ne 0 \right)$ are $\alpha$ and $\beta$ , then

\begin{align} & \alpha +\beta =\dfrac{-b}{a} \\ & \alpha \cdot \beta =\dfrac{c}{a} \\ \end{align}

Now, let us assume that the roots of the equations be:

${{x}^{2}}+2ax-{{b}^{2}}=0$ are x1 and x2.

$\text{ }{{x}^{2}}+2px-{{q}^{2}}=0$ are y1 and y2.

By substituting these in the above summation and product of the roots formula we get,
\begin{align} & {{x}_{1}}+{{x}_{2}}=-2a \\ & {{x}_{1}}\cdot {{x}_{2}}=-{{b}^{2}} \\ & {{y}_{1}}+{{y}_{2}}=-2p \\ & {{y}_{1}}\cdot {{y}_{2}}=-{{q}^{2}} \\ \end{align}

CIRCLE: Circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point in that plane is constant.

DIAMETER: A diameter of a circle is any straight line segment that passes through the centre of the circle and whose endpoints lie on the circle.

Equation of the circle, when the coordinates of endpoints of a diameter are
$\left( {{x}_{1}},{{y}_{1}} \right)\text{ }and\text{ }\left( {{x}_{2}},{{y}_{2}} \right)$:
$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$

Now, on multiplying the terms in this equation we get,

$\Rightarrow {{x}^{2}}-{{x}_{2}}x-{{x}_{1}}x+{{x}_{1}}\cdot {{x}_{2}}+{{y}^{2}}-{{y}_{2}}y-{{y}_{1}}y+{{y}_{1}}\cdot {{y}_{2}}=0$

$\Rightarrow {{x}^{2}}-\left( {{x}_{1}}+{{x}_{2}} \right)x+{{x}_{1}}\cdot {{x}_{2}}+{{y}^{2}}-\left( {{y}_{1}}+{{y}_{2}} \right)y+{{y}_{1}}\cdot {{y}_{2}}=0$

By substituting respective values in the above equation we get,

\begin{align} & \Rightarrow {{x}^{2}}-\left( -2a \right)x+\left( -{{b}^{2}} \right)+{{y}^{2}}-\left( -2p \right)y+\left( -{{q}^{2}} \right)=0 \\ & \Rightarrow {{x}^{2}}+2ax-{{b}^{2}}+{{y}^{2}}+2py-{{q}^{2}}=0 \\ \end{align}

Let us rearrange the terms in the above equation.

$\therefore {{x}^{2}}+{{y}^{2}}+2ax+2py-{{b}^{2}}-{{q}^{2}}=0$

Note: Instead of using the summation and product of the roots of the quadratic equation we can use direct formula and find the roots of the given quadratic equations which gives the coordinates of the points P and Q. After finding the roots of the equations we get coordinates of points P and Q. Now, as P and Q are the endpoints of the diameter we can find the midpoint which will be the centre of the circle. Then distance between this centre and one of the end points of the diameter gives the radius of the circle. By using the equation of the circle having centre (h, k) and radius a is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{a}^{2}}$. We can find the equation of the circle in this way also which gives the same result but a bit lengthy.