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  • The centre of the circle obtained by reflecting the circle <img alt="\mathrm{x^2+y^2=m^2+m+1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&

The centre of the circle obtained by reflecting the circle \mathrm{x^2+y^2=m^2+m+1} in the line \mathrm{y=m x+m}, is

 

Option: 1

\mathrm{\left(-\frac{2 m^2}{1+m^2},-\frac{2 m}{1+m^2}\right)}


Option: 2

\mathrm{\left(-\frac{2 m^2}{1+m^2}, \frac{2 m}{1+m^2}\right)}


Option: 3

\mathrm{\left(\frac{2 m}{1+m^2}, \frac{2 m^2}{1+m^2}\right)}


Option: 4

\mathrm{\left(\frac{2 m}{1+m^2},-\frac{2 m^2}{1+m^2}\right)}


Answers (1)

best_answer

Centre of the required circle is the reflection of the point (0, 0) in the line y = mx + m.

Let C (h, k) be the centre of the reflected circle 

\mathrm{\Rightarrow \frac{\mathrm{k}}{\mathrm{h}}=-\frac{1}{\mathrm{~m}}}              .... [1]

\mathrm{\text { and } \frac{\mathrm{k}}{2}=\mathrm{m} \frac{\mathrm{h}}{2}+\mathrm{m}}        ....[2]

\mathrm{\begin{aligned} & \Rightarrow k=m(-k m)+2 m \Rightarrow k=\frac{2 m}{1+m^2} \\ & \therefore C(h, k) \text { is }\left(-\frac{2 m^2}{1+m^2}, \frac{2 m}{1+m^2}\right) . \end{aligned}}

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Rishabh

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