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Question: A circle \(C\) whose radius is \[3\] ,touches externally the circle , \({x^2} + {y^2} + 2x - 4y - 4 ...

A circle CC whose radius is 33 ,touches externally the circle , x2+y2+2x4y4=0{x^2} + {y^2} + 2x - 4y - 4 = 0 at the point (2,2)\left( {2,2} \right) , then the length of the intercept cut by this circle CC, on the xx axis is equal to:
A) 232\sqrt 3
B) 323\sqrt 2
C) 5\sqrt 5
D) 252\sqrt 5

Explanation

Solution

Use the general formula for the circle to find the radius of the other circle. Observe that the two circles touch externally that means the sum of the radii is the distance between the centres of the circle. Use the formula to find the xx intercept for the circle CC.

Complete step-by-step answer:
The data given in the problem is,
The radius of the circle CC is 33.
x2+y2+2x4y4=0{x^2} + {y^2} + 2x - 4y - 4 = 0 at the point (2,2)\left( {2,2} \right) .
The general equation of a circle is x2+y2+2gx+2hy+c=0{x^2} + {y^2} + 2gx + 2hy + c = 0 .
The radius of the above circle is given by r=g2+f2cr = \sqrt {{g^2} + {f^2} - c} .
The centre of the circle is given by (g,h)( - g, - h).
Comparing the given equation with the general equation of the circle we observe that g=1,h=2,c=4g = 1,h = - 2,c = - 4.
Using the formula for the radius of the circle, we calculate the radius of the circle as follows:
r=12+(2)2(4)r = \sqrt {{1^2} + {{( - 2)}^2} - ( - 4)}
Therefore, r=3r = 3 .
Therefore, we observe that the radii of two circles are the same.
The centre of the other circle is (1,(2))=(1,2)\left( { - 1, - \left( { - 2} \right)} \right) = \left( { - 1,2} \right).
It is given that both circles touch each other externally at the point (2,2)\left( {2,2} \right).
Therefore, the point (2,2)\left( {2,2} \right) is the midpoint of the line segment joining both the centres.
Also, the length of the segment joining both centres is 3+3=63 + 3 = 6.
Let the centre of the circle CC be (a,b)\left( {a,b} \right).
We will use the midpoint formula to find the coordinates of the centre of the circle CC.
(1+a2,2+b2)=(2,2)\left( {\dfrac{{ - 1 + a}}{2},\dfrac{{2 + b}}{2}} \right) = \left( {2,2} \right)
Comparing both sides, we get (a,b)=(5,2)\left( {a,b} \right) = \left( {5,2} \right) .
Thus, the centre of the circle CC is (5,2)\left( {5,2} \right).
Equation of the circle CC is given as follows:
(x5)2+(y2)2=32{\left( {x - 5} \right)^2} + {\left( {y - 2} \right)^2} = {3^2}
Expanding the bracket and simplifying further we get x2+y210x4y+20=0{x^2} + {y^2} - 10x - 4y + 20 = 0.
Comparing with general equation of the circle we get g=5,h=2,c=20g = - 5,h = - 2,c = 20 .
The xx intercept is given by 2g2c2\sqrt {{g^2} - c} .
Therefore, xx intercept for circle CC is given as follows:
2g2c=2(5)2202\sqrt {{g^2} - c} = 2\sqrt {{{\left( { - 5} \right)}^2} - 20}
Therefore, on simplifying, we obtain that the intercept at the xx axis is 252\sqrt 5 .

So, the correct answer is “Option D”.

Note: We used the general equation of the circle and not the standard equation of the circle. The circles touch each other externally at a single point therefore, the sum of the radii is the distance between the centres of the circle. We calculate the xx intercept for circle CC by writing the equation of circle CC in general form.