Question
Question: The equation of the circle whose radius is 5 and which touches the circle \(x ^ { 2 } + y ^ { 2 } - ...
The equation of the circle whose radius is 5 and which touches the circle x2+y2−2x−4y−20=0 externally at the point (5, 5), is.
A
x2+y2−18x−16y−120=0
B
x2+y2−18x−16y+120=0
C
x2+y2+18x+16y−120=0
D
x2+y2+18x−16y+120=0
Answer
x2+y2−18x−16y+120=0
Explanation
Solution
Let the centre of the required circle be (x1,y1) and the centre of given circle is (1, 2). Since radii of both circles are same, therefore, point of contact (5, 5) is the mid point of the line joining the centres of both circles. Hence x1=9 and y1=8.Hence the required equation is(x−9)2+(y−8)2=25 ⇒x2+y2−18x−16y+120=0.
Trick : The point (5, 5) must satisfy the required circle. Hence the required equation is given by (2).