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Circles

SAT Math· difficulty 4/5

Where does the circle (x−3)2+(y−4)2=25(x-3)^2 + (y-4)^2 = 25 intersect the x-axis?

  • A

    At x=3±5x = 3 \pm 5

  • B

    At x=0x = 0 and x=6x = 6

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  • C

    At x=3x = 3 and x=−3x = -3

  • D

    At x=0x = 0 only

Explanation

Set y=0y = 0: (x−3)2+16=25(x-3)^2 + 16 = 25, so (x−3)2=9(x-3)^2 = 9, giving x−3=pm3x - 3 = \\pm 3, so x=0x = 0 or x=6x = 6.

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