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Question: Two circles \({x^2} + {y^2} - 2kx = 0\) and \({x^2} + {y^2} - 4x - 8y + 16 = 0\) touch each other ex...

Two circles x2+y2−2kx=0{x^2} + {y^2} - 2kx = 0 and x2+y2−4x−8y+16=0{x^2} + {y^2} - 4x - 8y + 16 = 0 touch each other externally. Then k is
(A) 44
(B) 11
(C) 22
(D) −4 - 4

Explanation

Solution

Hint: Here we know that if two circles touch each other externally then C1C2=R1+R2{C_1}{C_2} = {R_1} + {R_2}, where C1,C2{C_1},{C_2} represent the centre of two circles and R1,R2{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 x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0
Here we know that centre C=(−g,−f)C = ( - g, - f) and Radius R=g2+f2−cR = \sqrt {{g^2} + {f^2} - c}
Given circle are
x2+y2−2kx=0−−−−−−>(1){x^2} + {y^2} - 2kx = 0 - - - - - - > (1)
Centre C1=(k,0){C_1} = (k,0)
Radius R1=(−k)2=k{R_1} = \sqrt {{{( - k)}^2}} = k
x2+y2−4x−8y+16=0−−−−−−−−>(2){x^2} + {y^2} - 4x - 8y + 16 = 0 - - - - - - - - > (2)
Centre C2=(2,4){C_2} = (2,4)
R2=(−2)2+(−4)2−16=2{R_2} = \sqrt {{{( - 2)}^2} + {{( - 4)}^2} - 16} = 2
We know that if both circles touches externally then C1C2=R1+R2{C_1}{C_2} = {R_1} + {R_2}
⇒(k−2)2+(0−4)2=k+2\Rightarrow \sqrt {{{(k - 2)}^2} + {{(0 - 4)}^2}} = k + 2
Now let us square on both sides, then we get
⇒(k−2)2+(4)2=(k+2)2\Rightarrow {(k - 2)^2} + {(4)^2} = {(k + 2)^2}
⇒k2+4−4k+16=k2+4k+4\Rightarrow {k^2} + 4 - 4k + 16 = {k^2} + 4k + 4
⇒8k=16\Rightarrow 8k = 16
⇒k=2\Rightarrow k = 2

Thus, option C is the correct answer.

NOTE: If the two circles touch each other internally then we have:
⇒C1C2=∣R1−R2∣\Rightarrow {C_1}{C_2} = \left| {{R_1} - {R_2}} \right|
If the circles are intersecting at two different points then we have:
⇒∣R1−R2∣<C1C2<R1+R2\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:
⇒C1C2>R1+R2\Rightarrow {C_1}{C_2} > {R_1} + {R_2}