Find the equation of the circle whose radius is 5 and which touches the circle
x
2 + y2
-2x -4y -20 = 0 externally at the point (5,5).
Answers
Answer:
ANSWER
x
2
+y
2
−14x−10y−26=0−−−−−(1)
Comparing with standard equation x
2
+y
2
+2gx+2fy+c=0
g=−7 and f=−5, c=−26
Centre of (1)=(−g,−f)=(+7,+5)
Radius of (1)=
g
2
+f
2
−c
=
47+25+26
=
100
=10
Radius of circle where equation is to be found =5
Two circles touch at point (−1,3) and the circle lies inside of circle (1)
So, centre of internal circle is mid-point of (7,5) and (−1,3):(
2
7−1
,
2
5+3
)=(g,t)=(3,4)
∴ Required radius ⇒5=
9+16−c
Or, 25−c=25⇒c=−1
∴Required equation of internal circle:
x
2
+y
2
+6x+8y−1=
x+y4x 6y 12 0
Centre A is (2,3) and radius 5 = PA, B (h, k) is the
centre of the required circle of radius BP =3 which
touches the given circle internally at P (-1, - 1)
BA PA PB 53 =-2
Thus B divides PA in the ratio 3:2.
,k=3.2+2(-1)
3.2+2-1)
h =
3+2
3+2
5
7
5
Hence the required circle is
(x-4/5) +(y-7/5) 32
Or 5x+ 5y-8x-14y 32 0.
