Coordinate Geometry

JEE Main level, numerical answer, medium

3 of 5

Question 3 of 5

The radius of the circle given by equation x2+y2−6x+8y=0x^2 + y^2 - 6x + 8y = 0 is:

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Answer: 5

2g=−6  ⟹  g=−32g = -6 \implies g = -3, 2f=8  ⟹  f=42f = 8 \implies f = 4, c=0c = 0. Radius r=g2+f2−c=9+16=25=5r = \sqrt{g^2 + f^2 - c} = \sqrt{9 + 16} = \sqrt{25} = 5.

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