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  • Locus of the centre of the circle which touches the x-axis and also the circle  <img alt="\mathrm{(x-2)^2+(y-2)^2=4}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28x

Locus of the centre of the circle which touches the x-axis and also the circle  \mathrm{(x-2)^2+(y-2)^2=4}  externally is

Option: 1

\mathrm{(x-2)^2+(y-2)^2=16 }


Option: 2

\mathrm{(x-2)^2-8 y=0 }


Option: 3

\mathrm{(x-2)^2+\left(y-\frac{5}{2}\right)^2=10 }\mathrm{(x-2)^2+\left(y-\frac{5}{2}\right)^2=10 }


Option: 4

\mathrm{x^2-4 x+4=0}


Answers (1)

best_answer

The circle touches x-axis and  \mathrm{(x-2)^2+(y-2)^2=4}  externally, let its centre is (h,k) and radius |k|.

\mathrm{ \therefore \quad C_1 C_2=r_1+r_2 \Rightarrow \sqrt{(h-2)^2+(k-2)^2}=k+2 }
\mathrm{ \Rightarrow \quad(h-2)^2+(k-2)^2=(k+2)^2 }

Thus, locus is \mathrm{ (x-2)^2+(y-2)^2=(y+2)^2 }
\mathrm{ \Rightarrow(x-2)^2-8 y=0 }.

Posted by

seema garhwal

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