answer correctly got 10 points!?

Question

Give the standard form for the equation of the circle determined by the given condition.(choose any one)
1.Center(3,5),tangent to x-7y+2=0
2.through(1,2)with center at the point of intersection of x+3y+7=0 and 3x-2y-12=0.

Answer 1

1. r is the distance between the center poin (3,5)t to line x - 7y + 2 = 0

d = det((1*3)-(7*5)+2) / sqrt (1+49)
d= 30 / sqrt (50)
d= sqrt (900/50)
d = sqrt 18

so the circle equation is: (x-3)^2 + (y-5)^2 = 18


2. center point is when the two line intersects...
3x+9y+21=0
3x -2y -12=0
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11y = -33
y = - 3

x -9 + 7 = 0
x = 2

center point = (2,-3)

r = distance between the two points
r = sqrt ( (2-1)^2 + (2+3)^2)
r = sqrt (1 + 25)
r = sqrt (26)

the circle equation is... (x-2)^2 + (y+3) = 26

Answer 2

2).

x + 3y + 7 = 0
3x - 2y - 12 = 0

Using the substitute Method to solve this system of equations.

Solving for y

x = -3y - 7

Insert this into the x position in the following equation

3x -2y -12 = 0

3(-3 -7) - 2y - 12 = 0

- 9y -21 - 2y - 12 = 0

The distributive property

- 11y -33 = 0

Collecting terms

-11y -33 = 0
+ 33 + 33

Adding + 33 to both sides

- 11y = 33

-11y/-11 = 33/-11

Dividing -11 from both sides

y = -3

Insert the y value into the y position in the next equation.

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Solving fokr x

x = -3y - 7

x = -3(- 3) - 7

x = 9 - 7

x = 2

Insert the x value into the next equation

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Proof

x + 3y + 7 = 0

2 + 3(-3) + 7 = 0

2 + -9 + 7 = 0

0=0

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Proof

3x - 2y - 12 = 0

3(2) -2(-3) -12 = 0

6 + 6 - 12 = 0

12 -12 = 0

0 = 0

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The answer is:

x = 2

Y = -3

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The center Point

(2, -3)(1, 2)

Answer 3

The standard equation for a circle with rarius r and center at point (h,k) is

(x-h)^2 + (y-x)^2 = r^2

Problem 1 gives the center and says 'tangent to' which means it touches x-7y+2=0 at one (and only one) point. How far is it from (3,5) to x-7y+2=0? The distance (d) is
d=abs(3-35+2)/(√(50)) which is also the radius of the circle and r^2 = ((abs(3-35+2))/(√(50)))^2
So the final equation of the circle is
(x-3)^2 + (y-5)^2 = 18

For the second problem, solve the two linear equations with Gaussian elimination to get the point where they intersect. Now find the distance from that point to (1,2) and you have r. Then use the intersection as the origin in the equation for the circle and you're done.


DEoug

Answer 4

2. through (1,2) with center at the point intersection of x+3y+7=0 and 3x-2y-12=0

x+3y +7 = 0 --> x= -3y -7
3x -2y -12 = 0
3(-3y-7) -2y -12 =0
-9y -21 -2y -12 =0
11y = 33
y = 3
x= -9 -7 = -16

the point intersection is (-16,3) --> (a,b)
(1,2) - (-16,3) = (17,-1)
r² = 17² + (-1)² = 289+1 = 290

(x-a)² + (y-b)² = r²
(x-(-16))² + (y-3)² = 290
(x+16)² + (y-3)² = 290 or (x+16)² + (y-3)² -290 =0

Not satisfied?
(x² + 32x + 256) + (y² -6y + 9) - 290 = 0
x² + y² + 32x -6y -25 = 0

I hope I get 10 points, thx, c ya..

^_^

Answer 5

2.through(1,2)with center at the point of intersection of x+3y+7=0 and

Answer 6

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Answer 7

for second one : (x-2)*2 +(y+3)*2=26 ( the 2 s should be placed higher )

Answer 8

for 2nd q
x2 + y2- 4x +6y -13=0

Answer 9

Give me a break. And maybe 10 points? C'mon!

Answer 10

i don't really understand .....is this a trick question ?
or i didn't learn that lesson yet.(i'm 14)
or i don't know english that well.
anyway,thanks for the 10 points.