Question

find the equation of the line passes through the points(1,-2),(4,-3) and has its centre on line 3x+4y+10=0​

Answer 1

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