Question

#Quality Question
@Coordinate Geometry

Q》》
Equation of common tangent to circle
${x}^{2} + {y}^{2} - 6x = 0 \: \: and \: \: parabola \: \: \\ {y}^{2} = 4x \: \: \bf \: is$
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Answer 1

Answer:

${ \bf{ \color{violet}Let \: equation \: of \: tangent \: to \: the \: parabola \: \: \: \: { y }^{2} =4x  \: \: \: is}}</p><p>$

$y=mx+ \frac1m</p><p>$

$\sf\green{⇒{m }^{2} x−ym+1=0 \:  is \: tangent \: to  { x }^{2} +{y }^{2} −6x=0}</p><p></p><p></p><p>$

$\tt \huge \pink{⇒ \frac{∣3{m}^{2} +1∣}{ \sqrt{{m }^{4}+{m }^{2} }}=3}$

$\blue{m=± \frac{1}{ \sqrt31}}</p><p></p><p></p><p>$

$\color{cyan} \huge{⇒ tangent \: are \:  x+ \sqrt3y+3=0</p><p> \: \: \: \: \: \: \: and \: \: \: \: \: \: \: \: \: \:  x− \sqrt3y+3=0.}</p><p></p><p>$

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given equation of circle is

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 6x = 0$

So, its centre is evaluated as

$\rm \: = \: \bigg( - \dfrac{1}{2} \: coefficient \: of \: x, \: - \dfrac{1}{2} \: coefficient \: of \: y \bigg)$

$\rm \: = \: (3, \: 0)$

and

Radius is given by

$\rm :\longmapsto\: radius, \: r = \sqrt{ {g}^{2} + {f}^{2} - c}$

Here, g = - 3, f = 0, c = 0.

So, on substituting the values, we get

$\rm :\longmapsto\:radius, \: r = \sqrt{ {( - 3)}^{2} + {0}^{2} - 0}$

$\rm :\longmapsto\:radius, \: r = \sqrt{9}$

$\rm :\longmapsto\:radius, \: r = 3$

Also,

Given equation of Parabola is

$\rm :\longmapsto\: {y}^{2} = 4x$

We know,

If m is the slope of tangent to the parabola y² = 4ax, then equation of tangent is given by

$\rm :\longmapsto\:y = mx + \dfrac{a}{m}$

Now, given parabola is y² = 4x, so a = 1.

So, equation of tangent to parabola y² = 4x is

$\rm :\longmapsto\:y = mx + \dfrac{1}{m}$

$\rm :\longmapsto\:y = \dfrac{ {m}^{2}x + 1}{m}$

$\rm :\longmapsto\:my = x {m}^{2} + 1$

$\rm :\longmapsto\:x {m}^{2} - my + 1 = 0 - - - (1)$

For line (1) to be a common tangent to circle,

The perpendicular distance drawn from center (3, 0) on the line xm² - my + 1 = 0, must be equal to radius, i.e = 3.

So, using distance formula between point and line, we get

$\rm :\longmapsto\:3 = \dfrac{ |3 {m}^{2} - 0 + 1| }{ \sqrt{ {( {m}^{2} )}^{2} + {( - m)}^{2} } }$

$\rm :\longmapsto\:3 = \dfrac{ 3 {m}^{2} + 1}{ \sqrt{ {{m}^{4} } + { m}^{2} } }$

$\rm :\longmapsto\:3 \sqrt{ {m}^{4} + {m}^{2} } = {3m}^{2} + 1$

On squaring both sides, we get

$\rm :\longmapsto\:9( {m}^{4} + {m}^{2} ) = {9m}^{4} + 1 + 6 {m}^{2}$

$\rm :\longmapsto\:9{m}^{4} + 9{m}^{2} = {9m}^{4} + 1 + 6 {m}^{2}$

$\rm :\longmapsto\: {3m}^{2} = 1$

$\rm :\longmapsto\: {m}^{2} = \dfrac{1}{3}$

$\rm :\longmapsto\: {m}= \: \pm \: \dfrac{1}{ \sqrt{3} }$

So, Equation (1) can be rewritten as

$\rm :\longmapsto\:\dfrac{1}{3} x \: \pm \: \dfrac{1}{ \sqrt{3} } y + 1 = 0$

$\rm :\longmapsto\:\dfrac{1}{3} x \: \pm \: \dfrac{ \sqrt{3} }{3} y + \dfrac{1}{3} = 0$

$\rm :\longmapsto\:x \: \pm \: \sqrt{3}y + 3 = 0$

Hence,

The equation of common tangents to

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 6x = 0$

and

$\rm :\longmapsto\: {y}^{2} = 4x$

is

$\bf :\longmapsto\:x \: \pm \: \sqrt{3}y + 3 = 0$