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The circle in the figure above is centered at the origin, O (0, 0). Wh
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05 Feb 2024, 01:22
05 Feb 2024, 01:22
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The circle in the figure above is centered at the origin, O (0, 0). What is the slope of the line tangent to the circle at a point P (8, k)?
Indicate all possible values.
(A) 5/3
(B) 4/3
(C) 3/4
(D) -3/4
(E) -4/3
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Re: The circle in the figure above is centered at the origin, O (0, 0). Wh
[#permalink]
05 Feb 2024, 08:02
05 Feb 2024, 08:02
1
The answers are B and E.
The tangent will be perpendicular to the radius OP.
Distance formula:
--> OP = 10 = √(8 - 0)^2 + (k - 0)^2
Squaring both sides:
100 = 64 + k^2
--> k^2 = 36
--> k = 6 or -6
--> P = (8,6) or (8,-6)
Slope of OP = (y2 - y1) / (x2 - x1)
(x1,y1) = (0,0)
(x2,y2) = (8,6) or (8,-6)
--> m (OP) = 6/8 or -6/8 = 3/4 or -3/4
The tangent is perpendicular to OP, hence the slope will be the negative reciprocal of the slope of OP.
--> slope (tangent) = -4/3 or 4/3
The tangent will be perpendicular to the radius OP.
Distance formula:
--> OP = 10 = √(8 - 0)^2 + (k - 0)^2
Squaring both sides:
100 = 64 + k^2
--> k^2 = 36
--> k = 6 or -6
--> P = (8,6) or (8,-6)
Slope of OP = (y2 - y1) / (x2 - x1)
(x1,y1) = (0,0)
(x2,y2) = (8,6) or (8,-6)
--> m (OP) = 6/8 or -6/8 = 3/4 or -3/4
The tangent is perpendicular to OP, hence the slope will be the negative reciprocal of the slope of OP.
--> slope (tangent) = -4/3 or 4/3
gmatclubot
Re: The circle in the figure above is centered at the origin, O (0, 0). Wh [#permalink]
05 Feb 2024, 08:02
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