Question

A circle is described on a focal chord as diameter; if ‘m’ be the tangent of the inclination of the chord to the axis, prove that the equation to the circle is
${{x}^{2}}-2ax\left( 1+\dfrac{2}{{{m}^{2}}} \right)+{{y}^{2}}-\dfrac{4ay}{m}-3{{a}^{2}}=0$

Answer 1

Hint: Take the end points of chords as $\left( at_{1}^{2},2a{{t}_{1}} \right),\left( at_{2}^{2},2a{{t}_{2}} \right)$ then get the slope and use the formula $\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$ to get the equation of circle. After doing some further simplifications you will get the desired result.

Complete step-by-step solution -

We are given a focal chord as diameter; and ‘n’ be the target of the inclination of the chord to the axis.
Let assume that the end points of the focal chord be $P\left( at_{1}^{2},2a{{t}_{1}} \right)\text{ }$ and $Q\left( at_{2}^{2},2a{{t}_{2}} \right)$
If the points are $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right),$ then its slope will be $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
Now using the above formula, the slope PQ will be,
$m=\dfrac{2a\left( {{t}_{2}}-{{t}_{1}} \right)}{a\left( t_{2}^{2}-t_{1}^{2} \right)}$
$m=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{2}}-{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)}$
On further simplification we get,
$m=\dfrac{2}{{{t}_{1}}+{{t}_{2}}}$
So, we can write,
${{t}_{1}}+{{t}_{2}}=\dfrac{2}{m}................\left( i \right)$
If the endpoints of diameter of the circle is $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then the equation of circle will be,
$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$
Now we will apply the formula to get the equation of circle of points $\left( at_{1}^{2},2a{{t}_{1}} \right),\left( at_{2}^{2},2a{{t}_{2}} \right)$we get,
$\left( x-at_{1}^{2} \right)\left( x-at_{2}^{2} \right)+\left( y-2a{{t}_{1}} \right)\left( y-2a{{t}_{2}} \right)=0$
By further simplification we get,
${{x}^{2}}-ax\left( t_{1}^{2}+t_{2}^{2} \right)+{{a}^{2}}t_{1}^{2}t_{2}^{2}+{{y}^{2}}-2ay\left( {{t}_{1}}+{{t}_{2}} \right)+4{{a}^{2}}{{t}_{1}}{{t}_{2}}=0$
Now replacing $t_{1}^{2}+t_{2}^{2}$ by ${{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}-2{{t}_{1}}{{t}_{2}}$, so we can write the above equation as,
{{x}^{2}}-ax\left\\{ {{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}-2{{t}_{1}}{{t}_{2}} \right\\}+{{a}^{2}}t_{1}^{2}t_{2}^{2}+{{y}^{2}}-2ay\left( {{t}_{1}}+{{t}_{2}} \right)+4{{a}^{^{2}}}{{t}_{1}}{{t}_{2}}=0
Now as we know a certain condition that ${{t}_{1}}{{t}_{2}}=-1$ for focal chord, so the above equation can be written as,
{{x}^{2}}-ax\left\\{ {{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}+2 \right\\}+{{a}^{2}}+{{y}^{2}}-2ay\left( {{t}_{1}}+{{t}_{2}} \right)-4{{a}^{^{2}}}=0
On further simplification,
{{x}^{2}}-ax\left\\{ {{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}+2 \right\\}+{{y}^{2}}-2ay\left( {{t}_{1}}+{{t}_{2}} \right)-3{{a}^{^{2}}}=0
Now using equation (i) in the above equation, we get
{{x}^{2}}-ax\left\\{ \dfrac{4}{{{m}^{2}}}+2 \right\\}+{{y}^{2}}-2ay\left( \dfrac{2}{m} \right)-3{{a}^{2}}=0
On further simplification we can write it as,
${{x}^{2}}-2ax\left( 1+\dfrac{2}{{{m}^{2}}} \right)+{{y}^{2}}-\dfrac{4ay}{m}-3{{a}^{2}}=0$
Hence proved.

Note: Actually the $\left( at_{1}^{2},2a{{t}_{1}} \right)$ is a parametric point of parabola ${{y}^{2}}=4x$ which is often used in these kinds of problems. Also be careful about the calculation to avoid any errors.