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r1, r2 are th...

Question

$r_{1}, r_{2}$ are the radii of two non - intersecting circles A, B as centres. If P is the centre of AB then the perpendicular distance of P from their radical axis is

\frac{r_{1}^{2} + r_{2}^{2}}{2 A B}

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\frac{r_{2}^{2} - r_{1}^{2}}{2 A B}

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\frac{2 A B}{r_{1}^{2} - r_{2}^{2}}

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A B

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Solution

The correct option is B $\frac{r_{2}^{2} - r_{1}^{2}}{2 A B}$

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Similar questions

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle of radius R then

(a) $R_{1} + R_{2} = R$

(b) $R_{1} + R_{2} > R$

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

R_{1}

R_{2}

R_{1} + R_{2} = R

R_{1}^{2} + R_{2}^{2} = R^{2}

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R_{2}

\frac{{R_{1}}^{2}}{{R_{2}}^{2}} = \frac{9}{576}, \frac{R_{1}}{R_{2}} = ?

r_{1}

r_{2}

r

r = r_{1} + r_{2}

{r_{1}}^{2} + {r_{2}}^{2} = r^{2}

r_{1} + r_{2} < r

{r_{1}}^{2} + {r_{2}}^{2} < r^{2}

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