# What are the points of intersection of the circle x^2 - 6x + y^2 - 8y = 24 with the axes.

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### 2 Answers

The equation of the circle is x^2 - 6x + y^2 - 8y = 24.

Where the circle intersects the x-axis the y-coordinate is 0.

x^2 - 6x = 24

=> x^2 - 6x - 24 = 0

x1 = (6 + sqrt(36 + 96))/2

=> 3 + sqrt 33

x2 = 3 - sqrt 33

Where the circle intersects the y axis the x-coordinate is 0

y^2 - 8y = 24

=> y^2 - 8y - 24 = 0

y1 = 4 + sqrt 160/2 = 4 + sqrt 40

y2 = 4 - sqrt 40

**The x-intercepts of the circle are ****(3 + sqrt 33, 3 - sqrt 33)**** and the y-intercepts of the circle are ****(4+sqrt 40, 4 - sqrt 40)**

x^2-6x-y^2-8y=24

we substitute y=0 when circlre intersects the x_axis :

x^2-6x=24

x^2-6x-24=0

=> x1=(-b+sqrt(b^2-4(a)(c)))/2a=(6+sqrt132)/2=3+sqrt33

=>x2=(-b-sqrt(b^2-4(a)(c)))/2a=(6-sqrt132)/2=3-sqrt33

=>the x-interceptsof the circlr are (3+sqrt33,3-sqrt33)

and we need substitute x=0

when circle intersects the y_axis

=>y^2-8y-24=0

=>y1=4+2sqrt10

=>y2=4-2sqrt10

=>the y_intercecepts of the circle are (4+2 sqrt10,4-2 sqrt10)