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Question: Equation of a common tangent to the circle, \({x^2} + {y^2} - 6x = 0\) and the parabola, \({y^2} = 4...

Equation of a common tangent to the circle, x2+y2−6x=0{x^2} + {y^2} - 6x = 0 and the parabola, y2=4x{y^2} = 4x is
A)23y=12x+1 B)23y=−x−12 C)3y=x+3 D)3y=3x+1  A)2\sqrt {3y} = 12x + 1 \\\ B)2\sqrt {3y} = - x - 12 \\\ C)\sqrt 3 y = x + 3 \\\ D)\sqrt 3 y = 3x + 1 \\\

Explanation

Solution

Hint: Here, we proceed the solution by using the condition of tangent to the parabola and also by using the condition of tangent to the circle. Using these two conditions we will find the common tangent to the given circle and parabola.

Complete step-by-step answer:

Given, equation of circle of is x2+y2−6x=0{x^2} + {y^2} - 6x = 0 and parabola is y2=4x{y^2} = 4x
We know that equation of tangent to the parabola y2=4ax{y^2} = 4ax is y=mx+am−−−−−>(1)y = mx + \dfrac{a}{m} - - - - - > (1)
Here the given parabola is y2=4x{y^2} = 4x , where a=1
Now on substituting the value a=1 in equation (1)
The equation tangent to the given parabola y2=4x{y^2} = 4x will be
⇒y=mx+1m ⇒m2x−my+1=0  \Rightarrow y = mx + \dfrac{1}{m} \\\ \Rightarrow {m^2}x - my + 1 = 0 \\\
According to the question we can say that the equation of tangent is common for both circle and parabola.
Therefore, m2x−my+1=0{m^2}x - my + 1 = 0 is tangent to the circle x2+y2−6x=0{x^2} + {y^2} - 6x = 0
And we also know the condition that the line y=mx+cy = mx + c becomes tangent to the circle, when the perpendicular distance from centre to line is equal to the radius of the circle.

Perpendicular distance from a line to any point =∣ax+by+c∣a2+b2 = \dfrac{{|ax + by + c|}}{{\sqrt {{a^2} + {b^2}} }}
As it is already mentioned that the equation of tangent to parabola is tangent to the circle. Therefore by using the condition we say that the perpendicular distance from the centre of circle should be equal to zero.

⇒∣3m2+1∣m4+m2=3 ⇒m=±13  \Rightarrow \dfrac{{|3{m^2} + 1|}}{{\sqrt {{m^4} + {m^2}} }} = 3 \\\ \Rightarrow m = \pm \dfrac{1}{{\sqrt 3 }} \\\

Here we know that equation tangent is m2x−my+1=0{m^2}x - my + 1 = 0
To get equation of tangent let us substitute the ‘m’ value in the above equation, we get
⇒\Rightarrow x−3y+3=0x - \sqrt 3 y + 3 = 0 and x+3y+3=0x + \sqrt 3 y + 3 = 0
Here we have two equations of tangent as we have 2 m values.

NOTE: In this problem we used the condition for equation of tangent to the parabola and also condition for equation of tangent to the circle. By using these two conditions we got the m value which we substitute in m quadratic equation which tangent to both circle and parabola. This kind of problem instead of solving directly makes use of the conditions which are concept based and helpful to solve answers.