Question

find the equation of the circle which thought the point (1,-2) (4,-3) and has its create on the 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

⇒h2−2h+1+k2+4k+4=h2−8h+16+k2+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=10−53=547 or h=1547

The radius will be (1547−1)2+(5−3+2)2=(1532)2+(57)2=2251024+441=2251465

The circle equation is therefore (x−1547)2+(y+53)2=45293