Circle And System Of Circles Question 98
Question: The equation of the circle whose radius is 5 and which touches the circle $ x^{2}+y^{2}-2x-4y-20=0 $ externally at the point (5, 5), is
[Pb. CET 2003; IIT 1979]
Options:
A) $ x^{2}+y^{2}-18x-16y-120=0 $
B) $ x^{2}+y^{2}-18x-16y+120=0 $
C) $ x^{2}+y^{2}+18x+16y-120=0 $
D) $ x^{2}+y^{2}+18x-16y+120=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Let the centre of the required circle be $ (x_1,\ y_1) $ 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 $ x_1=9 $ and $ y_1=8 $ .
Hence the required equation is $ {{(x-9)}^{2}}+{{(y-8)}^{2}}=25 $
$ \Rightarrow x^{2}+y^{2}-18x-16y+120=0 $ . Trick : The point (5, 5) must satisfy the required circle.
Hence the required equation is given by (b).
BETA
