# If $y = mx$ be the equation of a chord of a circle whose radius is \[a\] , the origin of coordinates being one extremity of the chord and the axis of $x$ being a diameter of the circle, prove that the equation of a circle of which this chord is the diameter is \[\left( {1 + {m^2}} \right)\left( {{x^2} + {y^2}} \right) - 2a\left( {x + my} \right) = 0\].

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Hint-Equation of circle is \[{x^2}{\text{ + }}{y^2}{\text{ = }}{a^2}\] with centre at origin. The end points of the chord always lie on the circle and hence satisfy the circle equation.

According to question it is given that equation of chord is \[y = mx\]

So to solve this question first we assume the end of the chord be \[\left( {h,mh} \right)\] and the other end be \[\left( {0,0} \right)\]. These two end points of chords must lie on the circle.

Equation of circle is \[{\left( {x - a} \right)^2}{\text{ + }}{y^2}{\text{ = }}{a^2}\], and \[\left( {h,mh} \right)\] lies on the circle.

\[\therefore {\left( {h - a} \right)^2} + {m^2}{h^2} = {a^2}.....\left( {\text{i}} \right)\]

Now, expand the \[{\left( {h - a} \right)^2} = {h^2} + {a^2} - 2ha\] in the equation \[\left( {\text{i}} \right){\text{,}}\]

\[ \Rightarrow {h^2} - 2ah + {m^2}{h^2} = 0......\left( {{\text{ii}}} \right)\]

Take \[h\] common from above equation so that our equation simplifies further

\[

\Rightarrow h - 2a + {m^2}h = 0 \\

\Rightarrow h = \dfrac{{2a}}{{1 + {m^2}}} \\

\]

Now we got the value of $h$ which we can use to make the equation of the circle of which the chord is a diameter.

Therefore, the equation of circle of which chord $y = mx$ with endpoints \[\left( {0,0} \right){\text{ \& }}\left( {h,mh} \right)\] is the diameter will be

\[

(x - h)x + (y - mh)y = 0 \\

{x^2} + hx + {y^2} - mhy = 0 \\

{x^2} + {y^2} = h\left( {x + my} \right) \\

\]

Put $h$ value obtained as above

\[\left( {1 + {m^2}} \right)\left( {{x^2} + {y^2}} \right) = 2a(x + my)\]

Hence proved \[\left( {1 + {m^2}} \right)\left( {{x^2} + {y^2}} \right) = 2a(x + my) = 0\] is the equation of circle required.

Note-Whenever this type of question appears then always first write down the given things in the question. This is a very easy way to approach the question. Remember the standard equation of circle as mentioned in the solution which is \[{x^2}{\text{ + }}{y^2}{\text{ = }}{a^2}\]. The end points of a chord always lie on the circle hence, satisfy the circle equation.

According to question it is given that equation of chord is \[y = mx\]

So to solve this question first we assume the end of the chord be \[\left( {h,mh} \right)\] and the other end be \[\left( {0,0} \right)\]. These two end points of chords must lie on the circle.

Equation of circle is \[{\left( {x - a} \right)^2}{\text{ + }}{y^2}{\text{ = }}{a^2}\], and \[\left( {h,mh} \right)\] lies on the circle.

\[\therefore {\left( {h - a} \right)^2} + {m^2}{h^2} = {a^2}.....\left( {\text{i}} \right)\]

Now, expand the \[{\left( {h - a} \right)^2} = {h^2} + {a^2} - 2ha\] in the equation \[\left( {\text{i}} \right){\text{,}}\]

\[ \Rightarrow {h^2} - 2ah + {m^2}{h^2} = 0......\left( {{\text{ii}}} \right)\]

Take \[h\] common from above equation so that our equation simplifies further

\[

\Rightarrow h - 2a + {m^2}h = 0 \\

\Rightarrow h = \dfrac{{2a}}{{1 + {m^2}}} \\

\]

Now we got the value of $h$ which we can use to make the equation of the circle of which the chord is a diameter.

Therefore, the equation of circle of which chord $y = mx$ with endpoints \[\left( {0,0} \right){\text{ \& }}\left( {h,mh} \right)\] is the diameter will be

\[

(x - h)x + (y - mh)y = 0 \\

{x^2} + hx + {y^2} - mhy = 0 \\

{x^2} + {y^2} = h\left( {x + my} \right) \\

\]

Put $h$ value obtained as above

\[\left( {1 + {m^2}} \right)\left( {{x^2} + {y^2}} \right) = 2a(x + my)\]

Hence proved \[\left( {1 + {m^2}} \right)\left( {{x^2} + {y^2}} \right) = 2a(x + my) = 0\] is the equation of circle required.

Note-Whenever this type of question appears then always first write down the given things in the question. This is a very easy way to approach the question. Remember the standard equation of circle as mentioned in the solution which is \[{x^2}{\text{ + }}{y^2}{\text{ = }}{a^2}\]. The end points of a chord always lie on the circle hence, satisfy the circle equation.

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