TS ICET 23 May 2018 Shift 1 Paper

Question : 140 of 200

Marks: +1, -0

If the points A (l. 2), B(3, -4), C(5. -6) and D (ll, -8) lie on a circle, then the center of the circle is

Solution:

Let O(a, b) be the center of the circle.

\OA ^{2}=(a-1)^{2}+(b-2)^{2}

\OB^{2}=(a-3)^{2}+(b+4)^{2}

\OC ^{2}=(a-5)^{2}+(b+6)^{2}

\OD ^{2}=(a-11)^{2}+(b+8)^{2}

Since OA = OB = OC = OD
Equating

OA^2=OB^2

(a-1)^{2}+(b-2)^{2}=(a-3)^{2}+(b+4)^{2}

⇒4a = 12 b + 20

a = 3b + 5

→ (1)
Equating

OC^2 = OD^2

(a-5)^{2}+(b+6)^{2}=(a-11)^{2}+(b+8)^{2}

⇒12a=4b+124

3a=b+31

→ (2)
From (1) and (2), Solving for a and b
We get

a=11

and

b=2

Hence the center of the circle is (11, 2)

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