what is the equation of a circle passin through (6,20 and tangent to the line x-4y-15 at (3,-3)?

Answer 1

What is the equation of a circle passing through (6, 20) and tangent to the line x - 4y - 15 = 0 at (3,-3)?

The slope of the given line is m = ∆y/∆x = -1/-4 = 1/4. The center of the circle will be on a line perpendicular to the given line and passing thru the point (3,-3). The slope of the perpendicular line is the negative reciprocal of the slope of the given line.

m' = -1/m = -4

Use the point slope form to write the equation of the perpendicular line which contains the center of the circle.

y - -3 = -4(x - 3)
y + 3 = -4x + 12
4x + y = 9

The center of the circle is equidistant from the two given points on the circle (3,-3) and (6,20). It will lie on the perpendicular bisector of the two points. The midpoint M is:

M[(3+6)/2, (-3+20)/2] = M(9/2, 17/2).

The slope of the line segment is m1 = (20 - -3)/(6 - 3) = 23/3. The slope of the perpendicular bisector is m1' = -3/23.

The equation of the perpendicular bisector is:

y - 17/2 = (-3/23)(x - 9/2) = (-3/23)x + 27/46

Multiply thru by 46 to eliminated fractions.

46y - 391 = -6x + 27
6x + 46y = 418
3x + 23y = 209


The center of the circle is on the intersection of the two lines containing it.

4x + y = 9
3x + 23y = 209

Subtract three times the first equation from four times the second.

89y = 809
y = 809/89

Plug back into the first equation to solve for x.

4x + y = 9
4x + 809/89 = 9
4x = 9 - 809/89 = -8/89
x = -2/89

The center of the circle (h,k) = (-2/89, 809/89).

The radius is the distance from the center to one of the points on the circle. Let's choose (3, -3).

r² = (-2/89 - 3)² + (809/89 + 3)²
r² = (-269/89)² + (1076/809)² = 1,230,137 / 7,921
r = (√1,230,137) / 89 ≈ 12.461971

The equation of the circle is:

(x - h)² + (y - k)² = r²

(x + 2/89)² + (y - 809/89)² = 1,230,137/7,921

Answer 2

x - 4y - 15 = 0
4y = x - 15
y = (1/4)x - 15/4
perpendicular @ (3,-3):
y + 3 = - 4(x - 3)
A diameter of the circle has the equation
y = - 4x + 9 (1)
Now, a line from (3,-3) to (6,20) is a chord of the circle, with slope of 23/3 and midpoint (9/2,17/2)
The perpendicular bisector is
y - 17/2 = -3/23(x - 9/2)
46y - 391 = - 6x + 27
46y = - 6x + 418 (2)
Substituting (1) into (2),
46(- 4x + 9) = - 6x + 418
-184x + 414 = - 6x + 418
178x = - 4
x = - 4/178 = -2/89
y = 8/89 + 9 = 809/89
Center of the circle is (-2/89,809/89)
r^2 = (3 + 2/89)^2 + (- 3 - 809/89)^2
r^2 = (269/89)^2 + (-1076/89)^2
r^2 = (72361 + 1157776)/7921)
r^2 = 1230137/7921

(x + 2/89)^2 + (y - 809/89)^2 = 1230137/7921

(89x + 2)^2 + (89y - 809)^2 = 1,230,137
(x + 0.02247)^2 + (y - 9.089888)^2 = 155.3007

Answer 3

2x + 2y dy / dx = 0 dy / dx = - 2x / 2y dy / dx = - x / y 1 / 4 = - x / y y = - 4x x^2 + ( - 4x )^2 = r^2 17x^2 = r^2 17(6^2) = r^2 6 sqrt ( 17 ) = r so x^2 + y^2 = 612