The number of common tangents to the circles | vedclass

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(D) The given circles are ${x^2} + {y^2} - x = 0$ and ${x^2} + {y^2} + x = 0$.For the first circle ${x^2} + {y^2} - x = 0$,the centre ${C_1} = (\frac{

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$3$

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(D) The given circles are ${x^2} + {y^2} - x = 0$ and ${x^2} + {y^2} + x = 0$.For the first circle ${x^2} + {y^2} - x = 0$,the centre ${C_1} = (\frac{1}{2}, 0)$ and radius ${r_1} = \sqrt{(\frac{1}{2})^2 + 0^2 - 0} = \frac{1}{2}$.For the second circle ${x^2} + {y^2} + x = 0$,the centre ${C_2} = (-\frac{1}{2}, 0)$ and radius ${r_2} = \sqrt{(-\frac{1}{2})^2 + 0^2 - 0} = \frac{1}{2}$.The distance between the centres is ${C_1}{C_2} = \sqrt{(\frac{1}{2} - (-\frac{1}{2}))^2 + (0 - 0)^2} = \sqrt{1^2} = 1$.Since ${C_1}{C_2} = 1$ and ${r_1} + {r_2} = \frac{1}{2} + \frac{1}{2} = 1$,we have ${C_1}{C_2} = {r_1} + {r_2}$.This indicates that the two circles touch each other externally.When two circles touch each other externally,the number of common tangents is $3$.

The number of common tangents to the circles ${x^2} + {y^2} - x = 0$ and ${x^2} + {y^2} + x = 0$ is:

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