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Question

The equation of a common tangent to be circle $${x^2} + {y^2} = 16$$ and to the Ellipse $$\dfrac{{{x^2}}}{{49}} + \dfrac{{{y^2}}}{4} = 1$$ is 


A
y=x+45
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B
y=x+53
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C
11y=2x+4
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D
11y=2x+415
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Solution

The correct option is C $$\sqrt {11} y = 2x + 4\sqrt {15} $$

$${x^2} + {y^2} = 16$$

Equation of tangent to circle in slope form is

$$y = mx \pm \root r \of {1 + {m^2}} $$

$$y = mx \pm \root 4 \of {1 + {m^2}} $$   (1)

+Equation of tangent to euipse$${{{x^2}} \over {49}} + {{{y^2}} \over 4} \Rightarrow $$

Is $$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $$

$$y = mx \pm \sqrt {49{m^2} + 4} $$   (2)

(1) + (2) must be same so

$$4\sqrt {1 + {m^2}}  = \sqrt {49{m^2} + 4} $$

$$ \Rightarrow 16\left( {1 + {m^2}} \right) =  \pm \left( {49{m^2} + 4} \right)$$

$$ \Rightarrow 16 + 16{m^2} = 49{m^2} + 4$$

$$ \Rightarrow 33{m^2} = 12 \Rightarrow m =  \pm \sqrt {{4 \over {11}}}  = {2 \over {\sqrt {11} }}$$

$$y = \sqrt {{4 \over {11}}x}  + 4\sqrt {1 + {4 \over {11}}} $$

$$ = \sqrt {{4 \over {11}}x}  + {{4\sqrt {15} } \over {\sqrt {11} }}$$

$$\sqrt {11} y = 2x + 4\sqrt {15} $$


Mathematics

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