write the equation of the circle passing through...?

Question

a) passing through pts (8,-2), (6,2) and (3,-7)

b) passes through the origin, radius=10 and abcissa of the center is -6.

updating....

Answer 1

a)
It's best to draw a sketch on graph paper. The segment joining the points (6, 2) and (8, -2) has slope -2 and its midpoint is (7, 0), The perpendicular bisector of that segment has slope 1/2, and its equation is:
y = x/2 - 7/2

The segment joining the points (3, -7) and (8, -2) has slope 1 and its midpoint is (11/2, 9/2). The perpendicular bisector of that segment has slope -1 and its equation is:
y = -x + 1.

Set the two equations equal to each other:
x/2 - 7/2 = -x + 1
x - 7 = -2x + 2
3x = 9
x = 3
y = -x + 1
y = -3 + 1
y = -2
The center of the circle is (3, -2). That's 5 units from the point (8, -2), so the radius of the circle is 5
(x - h)^2 + (y - k)^2 = r^2
(x - 3)^2 + (y + 2)^2 = 25

b)
There are two such circles. First note the Pythagorean triple 6, 8, 10 puts the center 8 units from the x axis.
centers (-6, 8) and (-6, -8)
(x - h)^2 + (y - k)^2 = r^2
(x + 6)62 + (y - 8)^2 = 100 and
(x + 6)^2 + (y + 8)^2 = 100

Answer 2

general equation of a circle of radius r and center (xc, yc) is

(x -xc)^2 +(y -yc)^2 = r^2

a circle passing through (8,-2), (6,2) and (3,-7)
:

the distance from the center to any point on the circle = radius, hence'

distance from center to point (8, -2) is = distance from center to point (6,2).

r^2 = (xc -8)^2 +(yc +2)^2 = (xc -6)^2 +(yc -2)^2

expanding these , we get
xc^2 +64 -16xc +yc^2 +4 +4yc = xc^2 +36- 12xc +yc^2 +4 -4yc.

simplifying, we get
64 -16xc +4 +4yc = 36 -12xc +4 -4yc ;
68 -40 = -12xc -4yc +16xc -4yc
28 = 4xc -8yc
7 = xc -2yc --------> call this eq1
same thing will be done with points (6,2) and (3 , -7)

r^2 = (xc -6)^2 +(yc -2)^2 = (xc -3)^2 +(yc +7)^2;

36 -12xc +4 -4yc = 9 -6xc +49 +14yc
40 -58 = 6xc +18yc
-18 = 6xc +18yc
-3 = xc +3yc , or 3 = -xc -3yc -----> call this eq2

from eq1 and eq2, we get

7 = xc -2yc
3 = -xc -3yc

10 = -5yc ---> yc = -2
xc = 3

origin = (3, -2)

r^2 = (3 -8)^2 +(-2 +2)^2 = (3 -6)^2 +(-2 -2)^2 =
25

radius =5,

Equation of the circle passing throught the 3 points is

(x -3)^2 +(y +2)^2 =25;
x^2 +y^2 -6x +4y -12 =0

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b) passes through the origin, radius=10 and abcissa of the center is -6.


use the equation
(x -xc)^2 +(y -yc)^2 = r^2

and the points origin, center and radius...

Answer 3

enable the middle of the circle be (a , (4a-6)/3) we are able to write 2 equations of radius R^2 = (a-2)^2 + ((4a-6)/3 - 6)^2 and R^2 = (a-9)^2 + ((4a-6)/3 + 11)^2. you will come across a and relax.