Question

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

Answer 1

let equation of the tangent of circle be  y = mx + c

same line is tangent for y²=4x ... where a = 1 then tangent equation will be y = mx +1/m

( x − 3 )² + y² = 9  circle has radius as 3 and center at O( 3, 0)

perpendicular distance from center O to the tangent ( y = mx +1/m) will be Radius

so | [3m + 1/m ]/[√1+m²] | = 3

[ 3m² +1 ] = 3m [√1+m² ]  then squaring both sides

9m⁴ + 1 + 6m²  = 9m² [ 1+m² ]

1 = 3m²

m = 1/√3   or  -1/√3

tangent above X-axis will have positive slope then m = 1/√3

so Equation of tangent .... y = x/√3  + √3  

√3y = x + 3

Answer 2

Correct option is (C) \(\sqrt {3}y = x+3\)

\((x-3)^{2}+y^{2}=9 ; y^{2}=4 x\)

Tangent to \(y^{2}=4 x\)

\(y=m x+\frac{1}{m} \)

\( m^{2} x-m y+1=0 \) ....(1)

now centre of the circle \((3,0)\)

\(\left|\frac{3 m^{2}+1}{\sqrt{m^{4}+m^{2}}}\right|=3 \)

\( \left(3 m^{2}+1\right)^{2}=9\left(m^{4}+m^{2}\right)\)

\(\text {put } m^{2}=t\)

\( (3 t+1)^{2}=9\left(t^{2}+t\right) \)

\( 9 t^{2}+6 t+1=9 t^{2}+9 t \)

\( 3 t=1 \Rightarrow t=\frac{1}{3} \)

\(T: \frac{1}{3} x-\frac{1}{\sqrt{3}} y+1=0 \)

\( x-\sqrt{3} y+3-0 \Rightarrow \sqrt{3} y-x+3-0\).