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We have to write the equation of a family of circles passing through two given points $(x_1,y_1)$ & $(x_2, y_2)$

From these points , I only know one circle that is

$$(x_1-x)(x_2-x)+(y-y_1)(y-y_2)=0$$

But now how Can I write other equations .

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    make the ansatz $$(x-x_M)^2+(y-y_M)^2=R^2$$ – 2017-01-30
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    the middelpoints of these circles are situated on the perpendicular bisector – 2017-01-30
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    See http://math.stackexchange.com/questions/2120609/find-the-equation-of-the-circle-which-passes-through-the-origin-and-cuts-off/2120614#2120614 – 2017-01-30
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    Why does $S + \lambda L =0$ work? That is how we parametrise a typical circle passing through two points using the equation of the circle passing through them and the line passing through them. – 2017-01-30

2 Answers 2

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Family of circle passing through two given points $A(x_1, y_1)$ and $B(x_2, y_2)$.

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For more details see this.

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    Why we can do it like this, that is S+$\lambda$L=0 – 2017-01-30
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    You have to see 5th part. That is your answer. – 2017-01-30
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We know that $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) $ is the equation of the circle having the two points as it's diameter and that $\lambda \begin {vmatrix} x & y & 1\\ x_1 & y_1 & 1\\ x_2 & y_2 & 1\end {vmatrix} $ is the equation of a line passing through the two points. Thus, the required family of circles is $$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) +\lambda \begin {vmatrix} x & y & 1\\ x_1 & y_1 & 1\\ x_2 & y_2 & 1\end {vmatrix} =0 $$

Also see here. Hope it helps.

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    Is there any difference in your answer and the answer in the link I provided? – 2017-01-30
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    Its totally copy paste. – 2017-01-30
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    @Kanwaljit Singh Sir, I regret to say that you have provided a link which states the same thing that I had memorised for preparation. I am sorry that I didn't load your answer while writing mine. – 2017-01-30
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    Why we can do it like this, that is S+λL=0 – 2017-01-30
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    Oh its ok. I am taking my downvote back. Its completely ok. I have same mistake many times. But others cant understand. – 2017-01-30