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

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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