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If a circle be drawn, so as always to touch a given straight line and also a given circle, prove that the locus of its center is a parabola.

Last updated date: 13th Jul 2024
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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.

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.
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.