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Corresponding Points : Ellipse

If two sphere...

Question

If two spheres of radii $r_{1}$ and $r_{2}$ cut orthogonally, then the radius of the common circle is

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Solution

The correct option is D

In $Δ$ OPC, $c o s θ = \frac{r}{r_{1}}$
In $Δ$ O'PC, $s i n θ = \frac{r}{r_{2}}$
As, $c o s^{2} θ + s i n^{2} θ = 1$
$∴ {(\frac{r}{r_{1}})}^{2} + {(\frac{r}{r_{2}})}^{2} = 1 \Rightarrow r = \frac{r_{1} r_{2}}{\sqrt{r_{1}^{2} + r_{2}^{2}}}$ .

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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}

R_{1} + R_{2} = R

R_{1} + R_{2} < R

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

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