Question

6. Find the equation of the circle passing through the points

(iii) (20,3), (19,8) and (2,-9)
Also, find the centre and radius in each case.
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Answer 1

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Answer 2

x² + y² - 14x - 6y - 111 = 0

The center of the circle is (7,3) and the radius will be $\sqrt{(- 7)^{2} + (- 3)^{2} + 111} = \sqrt{169} = 13$ units.

Step-by-step explanation:

The general form of a circle equation is

x² + y² + 2gx + 2fy + c = 0, where the center of the circle is at (- g,- f) and radius is $\sqrt{g^{2} + f^{2} - c }$.

Now, all three points on the circle (20,3), (19,8) and (2,-9) will satisfy the equation of the circle.

So, 20² + 3² + 40g + 6f + c = 0

⇒ 40g + 6f + c = - 409 .......... (1)

And, 19² + 8² + 38g + 16f + c = 0

⇒ 38g + 16f + c = - 425 ............. (2)

Again, 2² + (- 9)² + 4g - 18f + c = 0

⇒ 4g - 18f + c = - 85 ................ (3)

Now, from equations (1) and (2) we get,

(40g - 38g) + (6f - 16f) = - 409 + 425

⇒ 2g - 10f = 16

⇒ g - 5f = 8 .............. (4)

Again, from equations (2) and (3) we get

(38g - 4g) + (16f + 18f) = -425 + 85

⇒ 34g + 34f = - 340

⇒ g + f = - 10 .......... (5)

Now, solving equations (4) and (5) we get,

f + 5f = - 10 - 8

⇒ 6f = - 18

⇒ f = - 3

Now, g = - 10 - (- 3) = - 7

Now, from equation (3) we get,

c = 18f - 4g - 85 =  18(- 3) - 4(- 7) - 85 = - 111

Therefore, the equation of the circle is

x² + y² - 14x - 6y - 111 = 0 (Answer)

The center of the circle is (7,3) and the radius will be $\sqrt{(- 7)^{2} + (- 3)^{2} + 111} = \sqrt{169} = 13$ units. (Answer)