Equation of a common tangent to the circle, x2+y2−6x=0 and the parabola, y2=4x, is
A
2√3y=−x−12
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B
2√3y=12x+1
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C
√3y=3x+1
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D
√3y=x+3
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Solution
The correct option is D√3y=x+3 The curves are y2=4x and x2+y2−6x=0⇒(x−3)2+y2=9 Equation of tangent to y2=4x is y=mx+1m...(1) Equation of tangent to the circle is y=m(x−3)+3√1+m2, where m is the slope. Since the two curves have common tangents, ⇒mx+1m=m(x−3)+3√1+m2 ⇒m=±1√3
Putting the value of m in (1), we get √3y=±(x+3) i.e., √3y=x+3