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Question

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\[{{(x-3)}^{2}}+{{y}^{2}}=1\] are:

(a) (0,0)

(b) (2,1)

(c) (2,1)

(d) (4,4)

Answer
Verified

Hint: To find the coordinates of the point on the parabola which has the least

distance from the circle, we will find the distance between the centre of the circle and any

point on the parabola taken in parametric form. Then, we will differentiate that distance and

equate it to zero to get the coordinates of the point which is at minimum distance from the

circle.

We have a parabola\[{{x}^{2}}=4y\] and a circle\[{{(x-3)}^{2}}+{{y}^{2}}=1\]. We want to find

the coordinates of the point on the parabola which has the least distance from the circle.

We know that any general point on the parabola of the

form\[{{x}^{2}}=4ay\]is\[(2at,a{{t}^{2}})\].

By substituting\[a=1\], we have the point on our parabola as\[(2t,{{t}^{2}})\].

We know that a circle of the form\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]has centre\[(h,k)\].

Thus the centre of the circle\[{{(x-3)}^{2}}+{{y}^{2}}=1\]is\[(3,0)\].

We know that the distance between any two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is \[\sqrt{{({x_1}-{x_2})}^2+{({y_1}-{y_2})}^2}\].

Substituting \[{{x}_{1}}=2t,{{y}_{1}}={{t}^{2}},{{x}_{2}}=3,{{y}_{2}}=0\] in the above formula,

we get the distance between the points to be\[\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}\].

We know that to find the minimum value of a function\[f(x)\], we have\[\dfrac{df(x)}{dx}=0\].

So, we will differentiate our function \[\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}\] with respect to\[t\].

Thus, we get

\[\Rightarrow \dfrac{d}{dt}\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}=\dfrac{1}{2\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}}\{2(2t-3)2+

4{{t}^{3}}\}=0\]

\[\begin{align}

& \Rightarrow 4(2t-3)+4{{t}^{3}}=0 \\

& \Rightarrow 8t-12+4{{t}^{3}}=0 \\

\end{align}\]

We will now factorize the above equation. Add and subtract\[4{{t}^{2}}\]from the equation.

Thus we have\[4{{t}^{3}}+8t-12=4{{t}^{3}}-4{{t}^{2}}+4{{t}^{2}}+8t-12\].

\[\begin{align}

& \Rightarrow 4{{t}^{3}}-4{{t}^{2}}+4{{t}^{2}}+8t-12=0 \\

& \Rightarrow 4{{t}^{2}}(t-1)+4({{t}^{2}}+2t-3)=0 \\

& \Rightarrow 4{{t}^{2}}(t-1)+4({{t}^{2}}+3t-t-3)=0 \\

& \Rightarrow 4{{t}^{2}}(t-1)+4(t(t+3)-1(t+3))=0 \\

& \Rightarrow 4{{t}^{2}}(t-1)+4(t-1)(t+3)=0 \\

& \Rightarrow 4(t-1)({{t}^{2}}+t+3)=0 \\

\end{align}\]

We observe that\[t=1\]satisfies the equation and the equation\[{{t}^{2}}+t+3\]has no real

roots.

Substituting the value of\[t\]in the point\[(2t,{{t}^{2}})\], we get\[(2,1)\].

So, the point on the parabola which is at minimum distance from the circle is (2,1).

Hence, the correct answer is \[(2,1)\].

Note: We can also solve this question by finding the foot of perpendicular from any general

point on parabola to the circle and then differentiating the distance between the foot of

perpendicular on circle and point on parabola and equating it to zero.

distance from the circle, we will find the distance between the centre of the circle and any

point on the parabola taken in parametric form. Then, we will differentiate that distance and

equate it to zero to get the coordinates of the point which is at minimum distance from the

circle.

We have a parabola\[{{x}^{2}}=4y\] and a circle\[{{(x-3)}^{2}}+{{y}^{2}}=1\]. We want to find

the coordinates of the point on the parabola which has the least distance from the circle.

We know that any general point on the parabola of the

form\[{{x}^{2}}=4ay\]is\[(2at,a{{t}^{2}})\].

By substituting\[a=1\], we have the point on our parabola as\[(2t,{{t}^{2}})\].

We know that a circle of the form\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]has centre\[(h,k)\].

Thus the centre of the circle\[{{(x-3)}^{2}}+{{y}^{2}}=1\]is\[(3,0)\].

We know that the distance between any two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is \[\sqrt{{({x_1}-{x_2})}^2+{({y_1}-{y_2})}^2}\].

Substituting \[{{x}_{1}}=2t,{{y}_{1}}={{t}^{2}},{{x}_{2}}=3,{{y}_{2}}=0\] in the above formula,

we get the distance between the points to be\[\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}\].

We know that to find the minimum value of a function\[f(x)\], we have\[\dfrac{df(x)}{dx}=0\].

So, we will differentiate our function \[\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}\] with respect to\[t\].

Thus, we get

\[\Rightarrow \dfrac{d}{dt}\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}=\dfrac{1}{2\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}}\{2(2t-3)2+

4{{t}^{3}}\}=0\]

\[\begin{align}

& \Rightarrow 4(2t-3)+4{{t}^{3}}=0 \\

& \Rightarrow 8t-12+4{{t}^{3}}=0 \\

\end{align}\]

We will now factorize the above equation. Add and subtract\[4{{t}^{2}}\]from the equation.

Thus we have\[4{{t}^{3}}+8t-12=4{{t}^{3}}-4{{t}^{2}}+4{{t}^{2}}+8t-12\].

\[\begin{align}

& \Rightarrow 4{{t}^{3}}-4{{t}^{2}}+4{{t}^{2}}+8t-12=0 \\

& \Rightarrow 4{{t}^{2}}(t-1)+4({{t}^{2}}+2t-3)=0 \\

& \Rightarrow 4{{t}^{2}}(t-1)+4({{t}^{2}}+3t-t-3)=0 \\

& \Rightarrow 4{{t}^{2}}(t-1)+4(t(t+3)-1(t+3))=0 \\

& \Rightarrow 4{{t}^{2}}(t-1)+4(t-1)(t+3)=0 \\

& \Rightarrow 4(t-1)({{t}^{2}}+t+3)=0 \\

\end{align}\]

We observe that\[t=1\]satisfies the equation and the equation\[{{t}^{2}}+t+3\]has no real

roots.

Substituting the value of\[t\]in the point\[(2t,{{t}^{2}})\], we get\[(2,1)\].

So, the point on the parabola which is at minimum distance from the circle is (2,1).

Hence, the correct answer is \[(2,1)\].

Note: We can also solve this question by finding the foot of perpendicular from any general

point on parabola to the circle and then differentiating the distance between the foot of

perpendicular on circle and point on parabola and equating it to zero.

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