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Consider the circle \mathrm{C: x^2+y^2=r^2} with centre O. A and B are collinear with O such that \mathrm{O A \cdot O B=r^2.} The number of circles passing through A and B, which are orthogonal to the circle C is

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

> 2


Answers (1)

best_answer

Let \mathrm{A \equiv(\alpha, 0)} and \mathrm{B \equiv\left(\frac{r^2}{\alpha}, 0\right)}

The circles through A and B are

\mathrm{ \begin{aligned} & (x-\alpha)\left(x-\frac{r^2}{\alpha}\right)+y^2+\lambda y=0 \\ & \text { or } x^2+y^2-\left(\alpha+\frac{r^2}{\alpha}\right) x+\lambda y+r^2=0, \lambda \in R \end{aligned} }

It is orthogonal to \mathrm{x^2+y^2-r^2=0} for all values of \lambda.

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Shailly goel

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