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find the diameter and center of the following circle

x^2+y^2-8x+1=34

x^2+y^2-8x+1=34

complete the square
x^2 - 8x + .... + y^2 = 33
x^2 - 8x + 16 + y^2 = 33+16
(x - 4)^2 + (y - 0)^2 = 49

so the centre is (4,0) and the radius = 7 ---> diameter is 14

Did you know?

Did you know that you can find the diameter and center of a circle by analyzing its equation? For example, in the equation x^2+y^2-8x+1=34, the center of the circle is determined by the terms involving x and y, which are (-8x) and (0y) respectively. The x-coordinate of the center can be found by taking the opposite sign of the coefficient of x, so in this case, the x-coordinate of the center is 8. Similarly, the y-coordinate of the center would be 0. Therefore, the center of this circle is (8, 0). To find the diameter, we need to solve for the radius first. By rearranging the equation and isolating the coefficients of the x^2 and y^2 terms, we get (x-4)^2 + y^2 = 34. Comparing this with the standard form of a circle's equation, (x-h)^2 + (y-k)^2 = r^2, we can see that the radius squared is 34. Therefore, the radius of this circle is the square root of 34. Finally, we can double the radius to find the diameter, which in this case would be 2 times the square root of 34.