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Mix Examples-Circle and System of Circles Questions in English

Class 11 Mathematics · 10-1.Circle and System of Circles · Mix Examples-Circle and System of Circles

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DifficultMCQ
The equation of the common tangent touching the circle $(x - 3)^2 + y^2 = 9$ and the parabola $y^2 = 4x$ above the $X$-axis is
A
$\sqrt{3}y = 3x + 1$
B
$\sqrt{3}y = -(x + 3)$
C
$\sqrt{3}y = x + 3$
D
$\sqrt{3}y = -(3x + 1)$

Solution

(C) Step $1$: The equation of a tangent to the parabola $y^2 = 4ax$ (where $a=1$) is $y = mx + \frac{a}{m}$, which is $y = mx + \frac{1}{m}$.
Step $2$: Rewrite this as $mx - y + \frac{1}{m} = 0$. This line is also tangent to the circle $(x - 3)^2 + y^2 = 3^2$, so the perpendicular distance from the center $(3, 0)$ to the line must equal the radius $3$.
Step $3$: Using the distance formula: $\frac{|m(3) - 0 + 1/m|}{\sqrt{m^2 + (-1)^2}} = 3$.
Step $4$: $|3m + 1/m| = 3\sqrt{m^2 + 1} \implies |\frac{3m^2 + 1}{m}| = 3\sqrt{m^2 + 1}$.
Step $5$: Squaring both sides: $\frac{(3m^2 + 1)^2}{m^2} = 9(m^2 + 1) \implies 9m^4 + 6m^2 + 1 = 9m^4 + 9m^2$.
Step $6$: $3m^2 = 1 \implies m^2 = 1/3$. Since the tangent is above the $X$-axis, we take $m = 1/\sqrt{3}$.
Step $7$: Substituting $m = 1/\sqrt{3}$ into $y = mx + 1/m$: $y = \frac{1}{\sqrt{3}}x + \sqrt{3} \implies \sqrt{3}y = x + 3$.

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