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What is the tangent line equation of the circle x2 plus y2 -6x plus 4y -7 equals 0 passing through the point 1 2?
Wiki User
∙ 7y ago
Differentiate the circle equation to find the slope at any given point.
x^2 + y^2 -6x + 4y -7 = 0
2x + 2y(dy/dx) - 6 + 4dy/dx =0
2x - 6 + (2y + 4) dy/dx =0
dy/dx = (6 - 2x) / (2y + +4)
At the point ( 1,2)
dy/dx = (6 - 2(1)) / (2(2) + 4)
dy/dx = 3 / 8 The slope /gradient of the tangent line.
At the point ( 1,2)
y - 2 = (3/8)(x - 1)
y - 2 = 3x/8 - 3/8
y = 3x/8 + 13/8
or
8y = 3x + 13
or
8y - 3x = 13
or
8y - 3x - 13 = 0
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lenpollock
Wiki User
∙ 7y agoThe tangent line equation touching the given circle works out as 2y = x+3 or as x-2y+3 = 0 in its general form
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