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Yogita Ingle 5 months ago

The base equation for a circle centered at (h,k) with radius r is:

(x - h)^{2}+ (y - k)^{2} = r^{2}

If the points are concyclic, they all lie on the graph for the same circle.

Thus, all points are solutions to a single equation of the circle

WE can plug each point into the base equation, and obtain a system of equationd

FOR POINT (-2,10)

Equation 1: (-2 - h)^{2} + (10 - k)^{2} = r^{2}

FOR POINT (1,11)

Equation 2: (1 - h)^{2}+ (11 - k)^{2} = r^{2}

FOR POINT (6,10)

Equation 3: (6 - h)^{2} + (10 - k)^{2} = r^{2}

FOR POINT (9,7)

Equation 4: (9 - h)^{2} + (7 - k)^{2} = r^{2}

From Equation 1 and 2 above, since they have 2 different expressions equivalent to r2, we can construct an equation involving variables x and h.

r^{2} = r^{2}

(-2 - h)^{2} + (10 - k)^{2} = (1 - h)^{2} + (11 - k)^{2}

(-2 - h)(-2 - h) + (10 - k)(10 - k) = (1 - h)(1 - h) + (11 - k)(11 - k)

(4 + 2h + 2h + h^{2}) + (100 -10k - 10k + k^{2}) = (1 - h - h + h^{2}) + (121 - 11k - 11k + k^{2})

h^{2} + 4h + 4 + k^{2} - 20k + 100 = h^{2}-2h + 1 + k^{2} -22k + 121

We can subtract h and k from both sides.

4h + 4 - 20k + 100 = -2h + 1 -22k + 122

4h -20k + 104 = -2h - 22k + 122

Move all h and k variables to the left ... move all constants to the right

6H + 2K = 18

SYSTEM EQUATION 1: 6h + 2k = 18

From equation 3 and 4 above, we go through the same process to obtain a second system equation.

r^{2} = r^{2}

(6 - h)^{2} + (10 - k)^{2} = (9 - h)^{2} + (7 - k)^{2}

(6 - h)(6 - h) + (10 - k)(10 - k) = (9 - h)(9 - h) + (7 - k)(7 - k)

36 - 6h - 6h + h^{2}+ 100 - 10k - 10k + k^{2} = 81 - 9h - 9h + h^{2} + 49 = 7k - 7k + k^{2}

h^{2} - 12h + 36 + k^{2} - 20k + 100 = h^{2} - 18h + 81 + k^{2}- 14k + 49

Simplify, and subtract h2 and k2from both sides

-12h - 20k + 136 = -18h - 14k + 130

Move the h and k variables to the left, and the constants to the right

-12h + 18h - 20k + 14k = 130 - 136

6h - 6k = -6

SYSTEM EQUATION 1: 6h - 6k = -6

6h +2k = 18

6h - 6k = -6

Multiply the second equation by -1, we have ... -6h + 6k = 6

Adding the 2 equations, we have

6h + 2k = 18

-6h + 6k = 6

8k =24

k = 3

Using substitution of k=3 into equation 6h + 2k = 18,

6h + 2(3) = 18

6h + 6 = 18

6h = 12

h = 2

Now, use one of the points on the circle ... (1, 11) and (h,k) ... (2,3), to find the radius

(1 - 2)^{2} + (11 - 3)^{2} = r^{2}

1 + 64 = r^{2}

65 = r^{2}

r = √65

Thus the equation of our circle is:

(x - 2)^{2} + (y - 3)^{2} = (√65)^{2}

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Tarun Kumar 8 months, 2 weeks ago

5/(5-r)=26/6+r+1
5(5-r)=2*6*5/(6-r+1)(6-r)(5-r)
(7-r)(6-r)=12
42-7r-6r-r2=12
-13r+r2+42-12=0
R2-13r+30=0
r2-3r-10r+30=0
r(r-3)-10(r-3)=0
=(r-3) (r-10)
=r-3=0 , r-10=0
r=3, r=10 (r is greater than n)

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Yogita Ingle 10 months, 3 weeks ago

Triangles are a three-sided polygon that consists of three edges and three vertices and the sum of internal angles of a triangle equal to 180^{o}.

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