The circles whose equations are Irl2 - 2r· (i + 2j) - 4 = 0 and...

The circles whose equations are Irl2 - 2r· (i + 2j) - 4 = 0 and Irl2 + 2r· (4i - 2j) + 6 = 0, intersect at the points A and B. Find

(a) the equation of the line AB, (b) the position vectors of A and B, (c) the equation of the circle through the origin and the points A and B, state its radius and the position vector of its centre.

Expert Solution

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Colby Messinger answered 1 year ago

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The circles whose equations are Irl2 - 2r· (i + 2j) - 4 = 0 and...

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