Question

The line $x+y=75$ intersects ${{y}^{2}}=4x$ at P and Q. Find the equation of circle as PQ diameter.

Answer 1

To find the equation of the circle, first we need to find the coordinates, P and Q. To find P and Q we should first substitute the value of any variable in the given parabolic equation. This will result in quadratic expression. Solve for the variable and then substitute both the values back into the line equation to get the values of the other variable too. Now use the formula for the circle when the diameter coordinates are given and then evaluate.

Complete step by step solution:
The given equations are, $x+y=75$ and ${{y}^{2}}=4x$
Let us now write the line equation in terms of x.
$\Rightarrow x+y=75$
$\Rightarrow x=75-y$
Now substitute this in the given parabolic equation, ${{y}^{2}}=4x$
$\Rightarrow {{y}^{2}}=4(75-y)$
Now evaluate this expression to get a quadratic expression.
$\Rightarrow {{y}^{2}}=300-75y$
Now write all the terms to solve the quadratic equation easily.
$\Rightarrow {{y}^{2}}+75y-300=0$
Now let us solve this expression.
This equation is in the form,$a{{x}^{2}}+bx+c=0$
Here,$a=1;b=75;c=-300$
Now put it in the formula,
$x=\left[ \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right]$
Substitute the values in the formula.
$\Rightarrow x=\left[ \dfrac{-\left( -1 \right)\pm \sqrt{{{\left( 75 \right)}^{2}}-4\times 1\times \left( -300 \right)}}{2\times 1} \right]$
Evaluate the expression under the square root first.
$\Rightarrow x=\left[ \dfrac{-\left( -1 \right)\pm \sqrt{5625+1200}}{2\times 1} \right]$
$\Rightarrow x=\left[ \dfrac{-\left( -1 \right)\pm \sqrt{6825}}{2\times 1} \right]$
Now simplify further.
$\Rightarrow x=\left[ \dfrac{1\pm \sqrt{6825}}{2\times 1} \right]$
$\Rightarrow x=\left[ \dfrac{1\pm 5\sqrt{273}}{2} \right]$
Hence the values of x are $\left[ \dfrac{1+5\sqrt{273}}{2} \right],\left[ \dfrac{1-5\sqrt{273}}{2} \right]$
Now substitute these values back in the line equation to get the values of y.
Since $x+y=75$
One value will be,
$\Rightarrow y=75-\dfrac{1+5\sqrt{273}}{2}$
Which can be further evaluated as,
$\Rightarrow y=\dfrac{149+5\sqrt{273}}{2}$
And the other one will be,
$\Rightarrow y=75-\dfrac{1-5\sqrt{273}}{2}$
Which can be further evaluated as,
$\Rightarrow y=\dfrac{149-5\sqrt{273}}{2}$
Therefore, the coordinates, P and Q are,
$P\left( \dfrac{1+5\sqrt{273}}{2},\dfrac{149+5\sqrt{273}}{2} \right)$
$Q\left( \dfrac{1-5\sqrt{273}}{2},\dfrac{149-5\sqrt{273}}{2} \right)$
The equation of a circle when the diameter’s coordinates are given is,
$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$
Hence on substituting in the formula we get,
$\Rightarrow \left( x-\dfrac{1+5\sqrt{273}}{2} \right)\left( x-\dfrac{1-5\sqrt{273}}{2} \right)+\left( y-\dfrac{149+5\sqrt{273}}{2} \right)\left( y-\dfrac{149-5\sqrt{273}}{2} \right)=0$
Upon further simplifying,
$\Rightarrow \left( \dfrac{2x-1+5\sqrt{273}}{2} \right)\left( \dfrac{2x-1-5\sqrt{273}}{2} \right)+\left( \dfrac{2y-149+5\sqrt{273}}{2} \right)\left( \dfrac{2y-149-5\sqrt{273}}{2} \right)=0$
$\Rightarrow \left( 2x-1+5\sqrt{273} \right)\left( 2x-1-5\sqrt{273} \right)+\left( 2y-149+5\sqrt{273} \right)\left( 2y-149-5\sqrt{273} \right)=0$
Since it will complicate more on expanding, we shall leave it here as it is.
Hence if the line $x+y=75$ intersects ${{y}^{2}}=4x$ at P and Q, then the equation of circle as PQ diameter is,
$\left( 2x-1+5\sqrt{273} \right)\left( 2x-1-5\sqrt{273} \right)+\left( 2y-149+5\sqrt{273} \right)\left( 2y-149-5\sqrt{273} \right)=0$

Note: Never forget to take two conditions whenever there is a $\pm$ symbol. The expression is written twice once with + and another time with - . Whenever there is a polynomial that is to be solved, the solution contains the roots of the expression. The number of roots is decided by the degree of the polynomial.