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Equation of circle: x^2 +y^2 -6x+4y+5 = 0

Completing the squares: (x-3)^2 +(y+2)^2 = 8

Radius of circle: square root of 8

Center of circle: (3, 2)

The tangent lines touches the circle on the x axis at: (1, 0) and (5, 0)

1st tangent equation: y = x-1

2nd tangent equation: y = -x+5

Note that the tangent line of a circle meets its radius at right angles

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6y ago

x2 + y2 - 6x + 4x + 5 = 0 => x2 + y2 - 2x + 5 = 0 => x2 - 2x + 1 + y2 = -5 + 1 = -4It is, therefore a circle with an imaginary radius of 2i units. Are you sure?

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Q: What are the tangent equations of the circle x2 plus y2 -6x plus 4x plus 5 equals 0 when it cuts through the x axis?
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