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Circle equationGiven the circle center (1,1) and the point on the circle (3,5) what is...

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colorcolour | Student, College Freshman | eNoter

Posted May 2, 2011 at 6:10 AM via web

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Circle equation

Given the circle center (1,1) and the point on the circle (3,5) what is the equation of the circle.

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giorgiana1976 | College Teacher | Valedictorian

Posted May 2, 2011 at 10:42 AM (Answer #2)

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We'll recall the general equation of the circle:

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

The coordinates of the center of circle are C(h ; k).

We know, from enunciation, that h = 1 and k = 1.

We'll replace them into the equation:

(x - 1)^2 + (y - 1)^2 = r^2

We'll determine the radius considering the constraint from enunciation that the circle is passing through the point (3,5).

If the circle is passing through the point (3,5), then the coordinates of the point are verifying the equation of the circle:

(3 - 1)^2 + (5 - 1)^2 = r^2

2^2 + 4^2 = r^2

4 + 16 = r^2

20 = r^2

r = 2sqrt5

Note: we'll reject th negative value of radius -2sqrt5.

The requested equation of the circle is: (x - 1)^2 + (y - 1)^2 = 20

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 3, 2011 at 12:45 PM (Answer #3)

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The center of the circle is (1,1). The point (3, 5) lies on the circle. The distance from (3,5) to (1,1) is the radius of the circle. This is equal to sqrt[(3 - 1)^2 + (5 - 1)^2] = sqrt [ 4 + 16] = sqrt 20

The equation of the circle is (x - 1)^2 + (y - 1)^2 = 20

=> x^2 - 2x + 1 + y^2 + 1 - 2y = 20

=> x^2 + y^2 - 2x - 2y - 18 = 0

The equation of the circle with center (1,1) and on which (3,5) lies is x^2 + y^2 - 2x - 2y - 18 = 0

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