5.
The diameters of a circle are along 2x+y-
7=0 and x+3y-11=0. Then the equation of
this circle which also passes through (5,7)
Answers
Answer:
x2+y2−4x−6y−12=0
Step-by-step explanation:
Given two diameters of circle
2x+y−7=0
x+3y−11=0
Given circle passes through point (5,7)
Lets assume circle equation is (x−h)2+(y−k)2=r2
(h,k)→ centre
r→ radius
By solving the diameters we get center as diameters passes through centre
2x+y−7=02x+6y−22=0______________−5y+15=0⇒y=3
2x+3−7=0
x−2
∴ centre (2,3)
point on circle =(5,7)
radius distance between centre and point
(x2−x1)2+(y2−y1)2
=(5−2)2+(7−3)2
=25
=5
By substituting centre (2,3) radius 5
in circle equation
(x−2)2+(y−3)2=52
x2+y2−4x−6y+13=25
x2+y2−4x−6y−12=0
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Answer:
hey mate
Step-by-step explanation:
Given two diameters of circle
2x+y−7=0
x+3y−11=0
Given circle passes through point (5,7)
Lets assume circle equation is (x−h)
2
+(y−k)
2
=r
2
(h,k)→ centre
r→ radius
By solving the diameters we get center as diameters passes through centre
2x+y−7=0
2x+6y−22=0
______________
−5y+15=0⇒y=3
2x+3−7=0
x−2
∴ centre (2,3)
point on circle =(5,7)
radius distance between centre and point
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(5−2)
2
+(7−3)
2
=
25
=5
By substituting centre (2,3) radius 5
in circle equation
(x−2)
2
+(y−3)
2
=5
2
x
2
+y
2
−4x−6y+13=25
x
2
+y
2
−4x−6y−12=0