Question 544496: Given that A(3,-3) is a point on the circle x^2 + y^2 -12x -2y+12 = 0.
a. Find the coordinates of B if AB is a diameter of the circle.
b. Find the value of k,k>0 if C(2,k) is a point on the circle
c. Show that triangleABC is an isosceles right-angled triangle
Answer by SwiftAlbatross(13)   (Show Source):
You can put this solution on YOUR website! First things first! Put the equation into standard form for a circle:
(x-H)² + (y-K)² = r² where (H, K) are the coordinates for the center of the circle and r is the radius. You can do this by completing the square for both x and y terms:
x² + y² - 12x - 2y + 12 = 0
x² + y² - 12x - 2y = -12
x² - 12x + y² - 2y = -12
To complete the square take half of the x-term and square it then add it to both sides of the equation. So take half of -12 which is -6, square it: -6² = 36 and add it to both sides.
x² - 12x + 36 + y² - 2y = -12 + 36
Do the same for y:
x² - 12x + 36 + y² -2y + 1 = -12 + 36 + 1
Now reduce each perfect square polynomial to squared form by remembering the FOIL rule for factoring:
(Ax + B)(Cx + D)
First + Outer + Inner + last
A²x² + ADx + BCx + D²
Now just work backwards and figure out what terms multiply to give you the trinomial you have. Let's start with the x :
x² - 12x + 36
Looks like 36 is perfect square of 6 and the negative sign in the middle term indicates subtract 6, so x - 6. Check it!
(x - 6)² = x² - 3x - 3x + 36 = x² - 6x + 36
Yes, it works! Now the y :
y² - 2y + 1
This looks good too since 1 is a perfect square:
(y - 1)² = y² - 1y - 1y + 1 = y² - 2y + 1
Perfect!
Now put it back into the equation:
x² - 12x + 36 + y² - 2y + 1 = -12 + 36 + 1
(x - 6)² + (y - 1)² = -12 + 36 + 1
Simplify the right:
(x - 6)² + (y - 1)² = 25
Beautiful! The equation of the circle! The center is (6,1) with radius 5. Now we can attack the parts in the question.
(a) First part asks us to find point B, knowing point A is (3,-3) and that AB is a diameter of the circle. Since we know the coordinates of the center and the fact that a diameter is just a line that goes through the center, we have 3 points on the line: A, B, and the center. We can just realize that the point B is equidistant and exactly opposite the point A with respect to the center. So let's find the difference between point A and the center, and move that distance to get to B:
center - A = (1 - -3) / (6 - 3) = (1 + 3) / (6 - 3) = 4 / 3
So let's move 4 y-spaces up and 3 x-spaces to the right to get to point B from the center:
x: 6 + 3 = 9
y: 1 + 4 = 5
So point B is (9,5).
(b) For (b) just plug point C back into the equation of the circle:
(x - 6)² + (y - 1)² = 25
(2 - 6)² + (k - 1)² = 25
-4² + (k - 1)² = 25
16 + (k - 1)² = 25
(k - 1)² = 9
Take the square root of both sides:
k - 1 = ±3
k - 1 = 3 OR k - 1 = -3
k = 4 OR k = -2
since k is above 0, we have
k = 4 and C(2,4)
(c) For part (c) just graph the points and convince yourself that this is an isosceles right triangle.
A(3,-3)
B(9,5)
C(2,4)
or you can use the distance formula and prove two sides are equidistant and pythagoras's theorem
AB = sqrt[(5 - -3)² + (9 - 3)²] = 10
BC = sqrt[(4 - 5)² + (2 - 9)²] = sqrt(50)
AC = sqrt[(2 - 3)² + (4 - -3)²] = sqrt(50)
a² + b² = c² where a and b are the same
a² + a² = c² --> 2a² = c² --> a = c / sqrt(2)
Look! The sides fit the description: sqrt(50) = 10 / sqrt(2)
sqrt(50) * sqrt(2) = 10
Square both sides
50 * 2 = 100
So it is an isosceles right triangle.
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