Answer:
Explanation:
To find the equation of a circle that passes through three given points, we can use the fact that the perpendicular bisectors of the chords joining the points intersect at the center of the circle.
Let's first find the midpoint and slope of the chords joining the three points:
The midpoint and slope of the chord joining (-1, 2) and (4, 2):
Midpoint: $((4 - 1)/2, (2 + 2)/2) = (3/2, 2)$
Slope: $(2 - 2)/(4 - (-1)) = 0$
The midpoint and slope of the chord joining (-1, 2) and (-3, 4):
Midpoint: $((-3 - 1)/2, (4 + 2)/2) = (-2, 3)$
Slope: $(4 - 2)/(-3 - (-1)) = 1/2$
The midpoint and slope of the chord joining (4, 2) and (-3, 4):
Midpoint: $((-3 + 4)/2, (4 + 2)/2) = (1/2, 3)$
Slope: $(4 - 2)/(-3 - 4) = -1/2$
Now we can find the equations of the perpendicular bisectors of these chords:
The equation of the perpendicular bisector of the chord joining (-1, 2) and (4, 2) is the horizontal line $y=2$.
The equation of the perpendicular bisector of the chord joining (-1, 2) and (-3, 4) is the line passing through the midpoint $(-2, 3)$ with slope $-2$:
$$y - 3 = -2(x + 2)$$
Simplifying, we get $y = -2x - 1$.
The equation of the perpendicular bisector of the chord joining (4, 2) and (-3, 4) is the line passing through the midpoint $(1/2, 3)$ with slope $2$:
$$y - 3 = 2(x - 1/2)$$
Simplifying, we get $y = 2x + 2$.
The center of the circle is the point where these perpendicular bisectors intersect. Solving the system of equations formed by setting any two of the perpendicular bisectors equal to each other, we get the center of the circle as $(1, 2)$.
Finally, the radius of the circle is the distance from the center to any of the three given points. We can use the distance formula to find that the radius is $\sqrt{10}$.
Putting it all together, the equation of the circle is:
$$(x - 1)^2 + (y - 2)^2 = 10$$
or expanding and simplifying:
$$x^2 + y^2 - 2x - 4y + 5 = 0$$
Therefore, the general equation for the circle that passes through the points (-1, 2), (4, 2), and (-3, 4) is $x^2 + y^2 - 2x - 4y + 5 = 0$.