Give the equation(s) of the circle that contains (-1,4) and (3,2) ; center on 3x - y + 3 = 0?

Answer 1

Equation of a circle with radius r and center (h,k) is:
(x - h)² + (y - k)² = r²
Solve equations:
(-1 - h)² + (4 - k)² = r²
(3 - h)² + (2 - k)² = r²
3*h - k + 3 = 0
(The first two equations express that points (-1,4) and (3,2) are on the circle. The third equation express that center of the circle is on the line 3x - y + 3 = 0.)
-

Answer 2

Let the coordinates of the centre be (h, k). Since the points (-1,4) and (3,2) lie on the circle, the distance of the centre from these two points is equal (each being equal to the radius of the circle. So, using distance formula we have
(h+1)^2+(k-4)^2 = (h-3)^2+(k-2)^2
=> h^2+2h+1+k^2-8k+16 = h^2-6h+9+k^2-4k+4
=> 8h-4k = -4
=> 2h-k = -1 (dividing both sides by 4) …..(i)
Also, the centre lies on the line 3x-y+3 = 0, which means
3h-k = -3 …...................(ii)
Subtracting equations (i) from (ii) we get
h = -2
Putting the value of h in equation (i) we have
-4-k = -1
=> -k = 3
=> k = -3
Hence the coordinates of the centre are (-2, -3). Using the distance formula we get radius as
r^2 = (-2-3)^2+(-3-2)^2 = 25+25 = 50

Now the equation of the circle is given as
(x-h)^2+(y-k)^2 = r^2
Put the values to get the required equation.