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Circles

SAT Math· difficulty 5/5

A circle has equation (x−3)2+(y−4)2=25(x-3)^2 + (y-4)^2 = 25. What is the equation of the tangent line to this circle at the point (0,0)(0, 0)?

  • A

    y=−34xy = -\frac{3}{4}x

  • B

    3x+4y=03x + 4y = 0

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  • C

    3x−4y=03x - 4y = 0

  • D

    4x+3y=04x + 3y = 0

Explanation

The radius from (3,4)(3,4) to (0,0)(0,0) has slope frac4−03−0=frac43\\frac{4-0}{3-0} = \\frac{4}{3}. The tangent is perpendicular, with slope −frac34-\\frac{3}{4}. Line through (0,0)(0,0): y=−frac34xy = -\\frac{3}{4}x, or 3x+4y=03x + 4y = 0.

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