Question

Two circles ${x^2} + {y^2} - 2kx = 0$ and ${x^2} + {y^2} - 4x - 8y + 16 = 0$ touch each other externally. Then k is
(A) $4$
(B) $1$
(C) $2$
(D) $ - 4$

Answer 1

Hint: Here we know that if two circles touch each other externally then ${C_1}{C_2} = {R_1} + {R_2}$, where ${C_1},{C_2}$ represent the centre of two circles and ${R_1},{R_2}$ represents the radius of circles. To get the value of k we have to substitute the values in the given condition.

Complete step by step answer:
The general equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$
Here we know that centre $C = ( - g, - f)$ and Radius $R = \sqrt {{g^2} + {f^2} - c} $
Given circle are
${x^2} + {y^2} - 2kx = 0 - - - - - - > (1)$
Centre ${C_1} = (k,0)$
Radius ${R_1} = \sqrt {{{( - k)}^2}} = k$
${x^2} + {y^2} - 4x - 8y + 16 = 0 - - - - - - - - > (2)$
Centre ${C_2} = (2,4)$
${R_2} = \sqrt {{{( - 2)}^2} + {{( - 4)}^2} - 16} = 2$
We know that if both circles touches externally then ${C_1}{C_2} = {R_1} + {R_2}$
$ \Rightarrow \sqrt {{{(k - 2)}^2} + {{(0 - 4)}^2}} = k + 2$
Now let us square on both sides, then we get
$ \Rightarrow {(k - 2)^2} + {(4)^2} = {(k + 2)^2}$
$ \Rightarrow {k^2} + 4 - 4k + 16 = {k^2} + 4k + 4$
$ \Rightarrow 8k = 16$
$ \Rightarrow k = 2$

Thus, option C is the correct answer.

NOTE: If the two circles touch each other internally then we have:
$\Rightarrow {C_1}{C_2} = \left| {{R_1} - {R_2}} \right|$
If the circles are intersecting at two different points then we have:
$\Rightarrow \left| {{R_1} - {R_2}} \right| < {C_1}{C_2} < {R_1} + {R_2}$
And if the circles are lying outside each other then:
$ \Rightarrow {C_1}{C_2} > {R_1} + {R_2}$