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  • The centre of a circle which is orthogonal to the circle <img alt="\mathrm{C}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}=25" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BC%7D_%7B1%7D%3A

The centre of a circle which is orthogonal to the circle \mathrm{C}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}=25 and \mathrm{\mathrm{C}_{2}:(\mathrm{x}-7)^{2}+(\mathrm{y}-5)^{2}=4} and which has the least radius is

Option: 1

\left(5, \frac{25}{7}\right)


Option: 2

(5,7)


Option: 3

(7,12)


Option: 4

none of these


Answers (1)

best_answer

\mathrm{C}_{1} \mathrm{C}_{2}=\sqrt{74}, \mathrm{r}_{1}+\mathrm{r}_{2}=7

\text{Since} \, \mathrm{C}_{1} \mathrm{C}_{2}>\mathrm{r}_{1}+\mathrm{r}_{2}

\Rightarrow  Circles do not intersect each other externally. The centre of the circle which is orthogonal to \mathrm{C}_{1} and \mathrm{C}_{2} and having least radius will lie on intersection of radical axis and the line joining the centre.

Equation of the radical axis: \mathrm{14 x+10 y=95}

Equation of the line joining centre \mathrm{\mathrm{y}=(5 / 7) \mathrm{x}}

\mathrm{Solving \, \mathrm{x}=\frac{95 \times 7}{148}}

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