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**Hint:**Take the given condition on plane and draw a line parallel to the given line at the distance equal to radius of given circle.

Given: A circle and a straight line.

We have to prove that locus of center of circle drawn touching given circle and straight line is parabola.

We have to prove that locus of center of circle drawn touching given circle and straight line is parabola.

**Complete step by step solution:**

Let the given line be \[h=0\] and given circle be \[{{C}_{1}}=0\] with center \[Q\] and radius \[{{R}_{o}}\].

Now, circle is drawn touching line \[h=0\] and circle \[{{C}_{1}}=0\] is \[{{C}_{2}}=0\] with center \[P\] and radius \[{{r}_{o}}\].

From diagram, we can see that,

\[PQ={{r}_{o}}+{{R}_{o}}.....\left( i \right)\]

As, \[{{r}_{o}}\] and \[PD\] are radius of circle \[{{C}_{2}}\],

We get, \[{{r}_{o}}=PD\]

Therefore, \[PQ=PD+{{R}_{o}}....\left( ii \right)\]

Construct a line \[{{h}_{1}}=0\] parallel to \[h=0\] at a distance \[{{R}_{o}}\] from \[h=0\] and extend \[PQ\] as shown.

Since \[h\] is parallel to \[{{h}_{1}}\],

Therefore, \[DR={{R}_{o}}...\left( iii \right)\]

Now, we know that

\[PQ={{r}_{o}}+{{R}_{o}}...\left( iv \right)\]

Therefore, \[PQ=PD+{{R}_{o}}\]….from equation \[\left( i \right)\]

Also, \[PR=PD+DR\text{ }\!\![\!\!\text { by diagram }\!\!]\!\!\text{ }\]

Therefore, \[PR={{r}_{o}}+{{R}_{o}}....\left( v \right)\left[ \text{From equation }\left( iii \right) \right]\]

Now, we know that any point on parabola has equal distances from its directrix and focus.

Similarly, here point \[P\] which is center of circle has equal distance \[\left[ {{r}_{o}}+{{R}_{o}} \right]\] from \[{{h}_{1}}=0\] and point \[Q\].

Therefore, here \[{{h}_{1}}\] is behaving as directrix and \[Q\] as a focus of parabola.

Now, circle is drawn touching line \[h=0\] and circle \[{{C}_{1}}=0\] is \[{{C}_{2}}=0\] with center \[P\] and radius \[{{r}_{o}}\].

From diagram, we can see that,

\[PQ={{r}_{o}}+{{R}_{o}}.....\left( i \right)\]

As, \[{{r}_{o}}\] and \[PD\] are radius of circle \[{{C}_{2}}\],

We get, \[{{r}_{o}}=PD\]

Therefore, \[PQ=PD+{{R}_{o}}....\left( ii \right)\]

Construct a line \[{{h}_{1}}=0\] parallel to \[h=0\] at a distance \[{{R}_{o}}\] from \[h=0\] and extend \[PQ\] as shown.

Since \[h\] is parallel to \[{{h}_{1}}\],

Therefore, \[DR={{R}_{o}}...\left( iii \right)\]

Now, we know that

\[PQ={{r}_{o}}+{{R}_{o}}...\left( iv \right)\]

Therefore, \[PQ=PD+{{R}_{o}}\]….from equation \[\left( i \right)\]

Also, \[PR=PD+DR\text{ }\!\![\!\!\text { by diagram }\!\!]\!\!\text{ }\]

Therefore, \[PR={{r}_{o}}+{{R}_{o}}....\left( v \right)\left[ \text{From equation }\left( iii \right) \right]\]

Now, we know that any point on parabola has equal distances from its directrix and focus.

Similarly, here point \[P\] which is center of circle has equal distance \[\left[ {{r}_{o}}+{{R}_{o}} \right]\] from \[{{h}_{1}}=0\] and point \[Q\].

Therefore, here \[{{h}_{1}}\] is behaving as directrix and \[Q\] as a focus of parabola.

**Hence, locus of \[P\] is parabola.****Note:**Students must understand the physical significances of coordinate geometry and importance of theory, specifically parabola in the given question.Recently Updated Pages

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