# How do i solve this equation of a circle problem?

Im not sure how to find the A and B coordinates from part a) Relevance
• (a) FInd the coordinates of A and B.

.....A = x-intercept = the point where y=0 so

..........0 = -3x+12

..........3x = 12

..........x = 4

....so A = (4,0)............ANS (see graph below)

.....B = y-intercept = the point where x=0 so

..........y = -3(0)+12

..........y = 12

.....so B = (0,12)...........ANS (see graph below)

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(b) Find the coordinates of the midpoint of AP

..............4+0....0+12

.....M = (------ , --------)

................2.........2

.....M = (2, 6)................ANS (see graph below)

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(c) Find the equation of the circle that passes through A(4,0), B(0,12) and O(0,0)

.....// Let M(2,6) be the center of the circle, then radius r =

.....r = AM = MB = √[(4-2)²+(12-6)²] = √40

.....// Verify that O lies on the circle:

.....// Check the distance of MO = r

.....MO = √[(2-0)²+(6-0)²] = √40

.....// The equation of the circle passing

.....// through A(4,0), B(0,12) and O(0,0)

.....// with center at M(2,6) is

.....(x-2)² + (y-6)² = 40..................ANS (see graph) • Let A be the x-intercept. y = 0

(0) = -3x + 12

x = 4

A(4, 0)

So B is the y-intercept. x = 0

y = -3(0) + 12

y = 12

B(0, 12)

Since ∠AOB is a right angle, AB must be a diameter of the circle.

(x - 4)(x - 0) + (y - 0)(y - 12) = 0

x² + y² - 4x - 12y = 0

As for the midpoint of AB, that is not needed, but it is (2, 6).

• Part (a) is just finding the intercepts and converting those to point coordinates.

For the x-intercept, let y=0 and solve for x:

0 = -3x + 12

3x = 12

x = 4

The x-intercept is 4, and the coordinates of that point are A(4, 0).

For the y-intercept, let x=0 and solve for y.

y = -3(0) + 12

y = 12

Or you could just "see" that the equation is in slope-intercept form.  Either y, they-intercept is 12 and the coordinates of that point are B(0, 12).

Part (b) involves remembering that the midpoint of segment AB has coordinates

M = (A+B)/2 = [(4, 0) + (0, 12)] / 2 = (4, 12)/2 = (2, 6).

Part (c) could be solved using distance formulas, but a little geometry makes this much easier.

>>> If you haven't already, make a drawing.  Really!  It helps a ton.

>>> Graph A, B, C and sketch the circle.

Since O and A are on the circle, the center of the circle is on the perpendicular bisector of OA.  Since OA is the horizontal segment from O(0,0) to A(4,0), the perpendicular bisector is clearly the vertical line x=2 (halfway between 0 and 4).

For the same reason, the center is also on the perpendicular bisector of chord OB, which is the horizontal line y = 6 (halfway between 0 and 12).

The intersection of the two lines is at C(2, 6) and that's your center.

Even simpler: If a right triangle is inscribed in a circle, the hypotenuse is a diameter of that circle, and the midpoint of that diamter is the center.  You already found that point in part (b).

Finally, the equation of the circle centered at C(2,6) is:

(x - 2)² + (y - 6)² = r²

...where r is the radius of the circle.  You can find r² as squared distance from O(0,0) to C(2,6):

r² = (2 - 0)² + (6 - 0)² = 4 + 36 = 40

...and the equation is:

(x - 2)² + (y - 6)² = 40

Obviously that works for point O.  Test for points A and B:

A: (4 - 2)² + (0 - 6)² = 4 + 36 = 40  ....  it works

B: (0 - 2)² + (12 - 6)² = 4 + 36 = 40  .... works here too

• a.

y = -3x + 12

When y = 0:

-3x + 12 = 0

x = 4

Hence, the coordinates of A = (4, 0)

When x = 0:

y = 12

Hence, the coordinates of B = (0, 12)

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b.

The coordinates of the midpoint of AB

= ((4 + 0)/2, (0 + 12)/2)

= (2, 6)

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c.

Let C(a, b) be the center of the circle.

(a - 0)² + (b - 0)² = (a - 4)² + (b - 0)²

a² = (a - 4)²

a² = a² - 8a + 16

-8a + 16 = 0

a = 2

(a - 0)² + (b - 0)² = (a - 0)² + (b - 12)²

b² = (b - 6)²

b² = b² - 12b + 36

-12b + 36 = 0

b = 3