Without drawing a diagram to scale, show that the circle ${x^2} + {\left( {y - 1} \right)^2} = 16$ lies completely inside the circle ${x^2} + {y^2} - x = 26$

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Hint: Here it is written that you have to answer without diagram but if it was not mentioned the best of answering these questions is through diagram. Here you have to answer by comparing distance between center and radius.

Complete step-by-step answer:
If the two circle ${C_1}{\text{ and }}{{\text{C}}_2}$ touch internally then ${C_1}{C_2} = {r_1} - {r_2}$. But if ${C_2}$ lies inside the circle
${C_1}$ then ${C_1}{C_2} < {r_1} - {r_2}$
Here ${C_1}\left( {\dfrac{1}{2},0} \right),{r_1} = \dfrac{1}{2}\sqrt {105} ,{C_2} = \left( {0,1} \right),{r_2} = 4$
Here $ {C_1}{C_2} $ is the distance between the centers of the given circles.
${C_1}{C_2} = \sqrt {\dfrac{1}{4} + 1} = \dfrac{1}{2}\sqrt 5 $
And ${r_1}{r_2} = \dfrac{1}{2}\sqrt {105} - 4$
Clearly ${C_1}{C_2} < {r_1} - {r_2}$. Hence circle ${C_2}$ lies completely inside the circle ${C_1}$.
Hence proved.

Note: Whenever you get this type of question the key concept of solving is you have to compare the distance between centers and difference of radii, if distance between centers is less than difference of radii then you can say one circle is completely inside the other circle.
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