Question

Determine the equation of family of circles passing through two given points $ A\left( {{x}_{1}},{{y}_{1}} \right) $ , $ B\left( {{x}_{2}},{{y}_{2}} \right) $ . If a fixed circle $ {{S}_{1}} $ cuts each member of the family in various chords, then prove that all these chords pass through a fixed point?

Answer 1

Hint: We start solving the problem by finding the equation of the circle with the given points $ A\left( {{x}_{1}},{{y}_{1}} \right) $ , $ B\left( {{x}_{2}},{{y}_{2}} \right) $ as ends of diameter. We then find the equation of the line segment passing through points $ A\left( {{x}_{1}},{{y}_{1}} \right) $ , $ B\left( {{x}_{2}},{{y}_{2}} \right) $ . We then find the equation of all the circles with line segment AB as chord and intersecting the circle with point A, B as ends of diameter using the fact that the equation of the circle intersecting the circle $ S=0 $ and having chord \[L=0\] is $ S+\lambda L=0 $ . We then assume equation of circle $ {{S}_{1}} $ and then find the equation of the chords formed due to intersection of $ {{S}_{1}} $ and obtained family of circles using the fact that the equation of the chord if the two circles $ {{S}_{a}} $ and $ {{S}_{b}} $ intersect each other is $ {{S}_{a}}-{{S}_{b}}=0 $ . We then compare them with the condition of concurrent lines to complete the obtained proof.

Complete step by step answer:
According to the problem, we are asked to find the equation of family of circles passing through two given points $ A\left( {{x}_{1}},{{y}_{1}} \right) $ , $ B\left( {{x}_{2}},{{y}_{2}} \right) $ . We then need to prove that all these chords pass through a fixed point if a fixed circle $ {{S}_{1}} $ cuts each member of the family in various chords.
Let us find the equation of the circle which has ends of diameter as $ A\left( {{x}_{1}},{{y}_{1}} \right) $ , $ B\left( {{x}_{2}},{{y}_{2}} \right) $ .
We know that the equation of the circle with ends of diameter $ \left( a,b \right) $ and $ \left( c,d \right) $ is $ \left( x-a \right)\left( x-c \right)+\left( y-b \right)\left( y-d \right)=0 $ .
So, we get the equation of the circle which has ends of diameter as $ A\left( {{x}_{1}},{{y}_{1}} \right) $ , $ B\left( {{x}_{2}},{{y}_{2}} \right) $ as $ \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0 $ .
\[\Rightarrow {{x}^{2}}-\left( {{x}_{1}}+{{x}_{2}} \right)x+{{x}_{1}}{{x}_{2}}+{{y}^{2}}-\left( {{y}_{1}}+{{y}_{2}} \right)y+{{y}_{1}}{{y}_{2}}=0\].
\[\Rightarrow {{x}^{2}}+{{y}^{2}}-\left( {{x}_{1}}+{{x}_{2}} \right)x-\left( {{y}_{1}}+{{y}_{2}} \right)y+{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}=0\] ---(1).
Let us find the equation of the line segment AB. We get it as $ y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\times \left( x-{{x}_{1}} \right) $ .
 $ \Rightarrow \left( y-{{y}_{1}} \right)\left( {{x}_{2}}-{{x}_{1}} \right)=\left( {{y}_{2}}-{{y}_{1}} \right)\times \left( x-{{x}_{1}} \right) $ .
\[\Rightarrow y\left( {{x}_{2}}-{{x}_{1}} \right)-{{y}_{1}}{{x}_{2}}+{{y}_{1}}{{x}_{1}}=x\left( {{y}_{2}}-{{y}_{1}} \right)-{{y}_{2}}{{x}_{1}}+{{y}_{1}}{{x}_{1}}\].
\[\Rightarrow x\left( {{y}_{2}}-{{y}_{1}} \right)-y\left( {{x}_{2}}-{{x}_{1}} \right)+{{y}_{1}}{{x}_{2}}-{{y}_{2}}{{x}_{1}}=0\] ---(2).
Now, we need to find the equation of the circles with the line segment joining points $ A\left( {{x}_{1}},{{y}_{1}} \right) $ , $ B\left( {{x}_{2}},{{y}_{2}} \right) $ as chord. But we can see that these circles intersect the obtained circle in equation (1).
We know that the equation of the circle intersecting the circle $ S=0 $ and having chord \[L=0\] is $ S+\lambda L=0 $ , where $ \lambda $ is an arbitrary parameter.
So, we get the equation of the family of circles as \[{{x}^{2}}+{{y}^{2}}-\left( {{x}_{1}}+{{x}_{2}} \right)x-\left( {{y}_{1}}+{{y}_{2}} \right)y+{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+\lambda \left( x\left( {{y}_{2}}-{{y}_{1}} \right)-y\left( {{x}_{2}}-{{x}_{1}} \right)+{{y}_{1}}{{x}_{2}}-{{y}_{2}}{{x}_{1}} \right)=0\] ---(3).
Now, let us assume the fixed circle $ {{S}_{1}}=0 $ be $ {{x}^{2}}+{{y}^{2}}+px+qy+r=0 $ . According to the problem, we are given that this circle cuts the obtained family of circles in equation (3) to form various chords.
We know that the equation of the chord if the two circles $ {{S}_{a}} $ and $ {{S}_{b}} $ intersect each other is $ {{S}_{a}}-{{S}_{b}}=0 $ .
Now, let us find the equation of all the chords formed which is as follows:
\[{{x}^{2}}+{{y}^{2}}-\left( {{x}_{1}}+{{x}_{2}} \right)x-\left( {{y}_{1}}+{{y}_{2}} \right)y+{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+\lambda \left( x\left( {{y}_{2}}-{{y}_{1}} \right)-y\left( {{x}_{2}}-{{x}_{1}} \right)+{{y}_{1}}{{x}_{2}}-{{y}_{2}}{{x}_{1}} \right)-\left( {{x}^{2}}+{{y}^{2}}+px+qy+r \right)=0\].
 $ \Rightarrow -\left( {{x}_{1}}+{{x}_{2}}+p \right)x-\left( {{y}_{1}}+{{y}_{2}}+q \right)y+\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}-r \right)+\lambda \left( x\left( {{y}_{2}}-{{y}_{1}} \right)-y\left( {{x}_{2}}-{{x}_{1}} \right)+\left( {{y}_{1}}{{x}_{2}}-{{y}_{2}}{{x}_{1}} \right) \right)=0 $ .
We can see that this equation of all the chords resembles the equation of concurrent lines. We know that concurrent lines pass through a fixed point.
 $\therefore$ We have proved that the obtained chords pass through a fixed point.

Note:
We should know that in order to have a unique circle, we need three non-collinear points. Since we have only two points, we get the infinite number of circles with line segment AB as diameter and as chord which is the reason, we followed this method. We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully in order to avoid confusion.