find the equation of the line passes through the points(1,-2),(4,-3) and has its centre on line 3x+4y+10=0
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Answer:
Let the center of the circle be (h,k)
Since the center is equidistant from all the points on the circle,
(h−1)
2
+(k+2)
2
=(h−4)
2
+(k+3)
2
⇒h
2
−2h+1+k
2
+4k+4=h
2
−8h+16+k
2
+6k+9
⇒6h−2k−20=0 or 3h−k−10=0 ........(1)
The center also satisfies 3h+4k−7=0
Finding the intersection of these lines, we get
3h−k−10 = 3h+4k−7
Put value in equation (1)
5k+3=0 or k=
5
−3
∴3h=10−
5
3
=
5
47
or h=
15
47
The radius will be
(
15
47
−1)
2
+(
5
−3
+2)
2
=
(
15
32
)
2
+(
5
7
)
2
=
225
1024+441
=
225
1465
The circle equation is therefore (x−
15
47
)
2
+(y+
5
3
)
2
=
45
293
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