Find the centre of the circle that passes through the points p(-9,5), q(1,5), and r(-2,-2)
consider the two chords PQ and PR
the centre must be the intersection of the right-bisector of these two chords.
midpoint of PQ= (-4,5)
slope of PQ = 0
equation of right-bisector of PQ is x = -4
(no real work required for this one, notice that PQ was a horizontal line, so the right-bisector must be the vertical line running through its midpoint)
midpoint of PR = (-11/2 , 3/2)
slope of PR = -7/7 = -1
so slope of perpendicular = +1
equation: x - y = C
but (-11/2 , 3/2) is on it
-11/2 -3/2 =c = -7
x - y = -7
sub in x = -4
-4 - y = -7
-y = -3
y = 3
the centre is (-4,3)
method2:
let the centre be (a,b)
the distance from (a,b) to each of those points must be the same ..., so
√((a+9)^2 + (b-5)^2) = √((a-1)^2 + (b-5)^2)
a^2 + 18a + 81 + b^2 - 10b + 25 = a^2 - 2a + 1 + b^2 -10b + 25
20a = -80
a = -4 , as before
√( (a+9)^2 + (b-5)^2) = √(a+2)^2 + (b+2)^2)
a^2 + 18a + 81 + b^2 - 10b + 25 = a^2 + 4a + 4 + b^2 + 4b + 4
14a -14b = -98
a - b = -7
but a=-4
-4 - b = -7
b = 3
then centre is (-4,3) , as before
To find the center of a circle passing through three given points, you can use the concept of the circumcenter. The circumcenter is the point where the perpendicular bisectors of the sides of a triangle intersect.
Here's how you can find the center of the circle that passes through the points P(-9,5), Q(1,5), and R(-2,-2):
1. Find the equations of the perpendicular bisectors of the sides.
- To find the equation of the perpendicular bisector of a side, you need the midpoint of that side and the negative reciprocal of its slope.
2. Find the intersection point of two of the perpendicular bisectors.
- Select any two sides of the triangle and find the equations of their perpendicular bisectors.
- Solve the two equations for the coordinates of the intersection point.
3. Repeat step 2 for another pair of perpendicular bisectors.
- Select another pair of sides and find their corresponding perpendicular bisectors.
- Find their intersection point.
4. The intersection point obtained in steps 2 and 3 will be the center of the circle that passes through the three given points.
Let's go through the process step by step:
1. Find the midpoint of each side:
- The midpoint of PQ is [(Px + Qx)/2 , (Py + Qy)/2]
= [(-9 + 1)/2 , (5 + 5)/2]
= [-4, 5]
- The midpoint of QR is [(Qx + Rx)/2 , (Qy + Ry)/2]
= [(1 + (-2))/2 , (5 + (-2))/2]
= [-0.5, 1.5]
- The midpoint of RP is [(Rx + Px)/2 , (Ry + Py)/2]
= [((-2) + (-9))/2 , ((-2) + 5)/2]
= [-5.5, 1.5]
2. Find the slope and negative reciprocal of each side:
- For PQ, the slope is (Qy - Py) / (Qx - Px)
= (5 - 5) / (1 - (-9))
= 0 / 10
= 0
So, the negative reciprocal is undefined.
- For QR, the slope is (Ry - Qy) / (Rx - Qx)
= ((-2) - 5) / ((-2) - 1)
= (-7) / (-3)
= 7/3
So, the negative reciprocal is -3/7.
- For RP, the slope is (Py - Ry) / (Px - Rx)
= (5 - (-2)) / (-9 - (-2))
= 7 / (-7)
= -1
So, the negative reciprocal is 1.
3. Find the equations of the perpendicular bisectors:
- For PQ, since the slope is undefined, the equation of the perpendicular bisector passing through the midpoint (-4, 5) is x = -4.
- For QR, the equation of the perpendicular bisector passing through the midpoint (-0.5, 1.5) can be found by using the point-slope form:
y - y1 = m(x - x1), where m is the negative reciprocal of the slope and (x1, y1) is the midpoint.
Therefore, y - 1.5 = (-3/7)(x + 0.5)
Simplifying, 7y - 10.5 = -3x - 1.5
Rearranging the equation, 3x + 7y = -9
- For RP, the equation of the perpendicular bisector passing through the midpoint (-5.5, 1.5) can be found similarly:
y - 1.5 = (1)(x + 5.5)
Simplifying, y - 1.5 = x + 5.5
Rearranging, x - y = -4
4. Now, we need to find the intersection point of any two perpendicular bisectors. We'll find the intersection point of the lines x = -4 and 3x + 7y = -9:
- Substituting the equation x = -4 into 3x + 7y = -9 gives 3(-4) + 7y = -9
Simplifying, -12 + 7y = -9
Rearranging, 7y = -9 + 12
Further simplification, 7y = 3
Solving for y, y = 3/7
Since the x-coordinate of the intersection point is fixed at -4, the center of the circle is (-4, 3/7).
So, the center of the circle that passes through the points P(-9,5), Q(1,5), and R(-2,-2) is (-4, 3/7).