Find the equation ofa circle passing through (1,2),(3,_4),(5,_6)
Answers
Step-by-step explanation:
Let the equation of the required circle be x
2
+y
2
+2gx+2fy+c=0
\textsf{since the given circle passes through (1,2), (3,-4) and (5,-6), we have}since the given circle passes through (1,2), (3,-4) and (5,-6), we have
\mathsf{1^2+2^2+2g(1)+2f(2)+c=0}1
2
+2
2
+2g(1)+2f(2)+c=0
\implies\mathsf{2g+4f+c=-5}⟹2g+4f+c=−5 ........(1)
\mathsf{3^2+(-4)^2+2g(3)+2f(-4)+c=0}3
2
+(−4)
2
+2g(3)+2f(−4)+c=0
\implies\mathsf{6g-8f+c=-25}⟹6g−8f+c=−25 ........(2)
\mathsf{5^2+(-6)^2+2g(5)+2f(-6)+c=0}5
2
+(−6)
2
+2g(5)+2f(−6)+c=0
\implies\mathsf{10g-12f+c=-61}⟹10g−12f+c=−61 ........(3)
\mathsf{(2)-(1), gives}(2)−(1),gives
\mathsf{6g-8f+c=-25}6g−8f+c=−25 ........(2)
\mathsf{2g+4f+c=-5}2g+4f+c=−5 ........(1)
\mathsf{4g-12f=-20}4g−12f=−20
\mathsf{g-3f=-5}g−3f=−5 ........(4)
\mathsf{(3)-(2), gives}(3)−(2),gives
\mathsf{10g-124f+c=-61}10g−124f+c=−61 ......(3)
\mathsf{6g-8f+c=-25}6g−8f+c=−25 ...........(2)
\mathsf{4g-4f=-36}4g−4f=−36
\mathsf{g-f=-9}g−f=−9 ...........(5)
\mathsf{g-3f=-5}g−3f=−5 ........(4)
\mathsf{(5)-(4), gives}(5)−(4),gives
\mathsf{2f=-4}2f=−4
\implies\mathsf{f=-2}⟹f=−2
\textsf{put f=-2 in (5), we get}put f=-2 in (5), we get
\mathsf{g+2=-9}g+2=−9
\implies\mathsf{g=-11}⟹g=−11
\textsf{put g=-11 and f=-2 in (1), we get}put g=-11 and f=-2 in (1), we get
\mathsf{2(-11)+4(-2)+c=-5}2(−11)+4(−2)+c=−5
\mathsf{-22-8+c=-5}−22−8+c=−5
\mathsf{c=25}c=25
\implies\mathsf{c=25}⟹c=25
\textsf{The equation of the required circle is}The equation of the required circle is
\mathsf{x^2+y^2+2(-11)x+2(-2)y+25=0}x
2
+y
2
+2(−11)x+2(−2)y+25=0
\implies\boxed{\mathsf{x^2+y^2-22x-4y+25=0}}⟹
x
2
+y
2
−22x−4y+25=0
\textsf{Its centre is (-g,-f)=(11,2)}Its centre is (-g,-f)=(11,2)
\textsf{Radius=$\sqrt{g^2+f^2-c}$}Radius=
g
2
+f
2
−c
\textsf{Radius=$\sqrt{121+4-25}$}Radius=
121+4−25
\textsf{Radius=$\sqrt{100}$}Radius=
100
\implies\textsf{Radius=10}⟹Radius=10