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write the standard form of the equation and the general form of the equ…

Question

write the standard form of the equation and the general form of the equation of the circle with radius r and center (h,k). then graph the circle.
r = 5; (h,k)=(4, - 3)
the standard form of the equation of this circle is (x - 4)^2+(y + 3)^2 = 25.
the general form of the equation of this circle is x^2 + y^2-2x = 0. (simplify your answer.)

Explanation:

Step1: Recall circle - standard - form formula

The standard - form of the equation of a circle with center \((h,k)\) and radius \(r\) is \((x - h)^2+(y - k)^2=r^2\).

Step2: Substitute given values

Given \(h = 4\), \(k=-3\), and \(r = 5\), we substitute into the formula: \((x - 4)^2+(y-(-3))^2=5^2\), which simplifies to \((x - 4)^2+(y + 3)^2=25\).

Step3: Expand to get general - form

Expand \((x - 4)^2+(y + 3)^2=25\).
\((x - 4)^2=x^{2}-8x + 16\) and \((y + 3)^2=y^{2}+6y+9\).
So \(x^{2}-8x + 16+y^{2}+6y + 9=25\).
Combine like - terms: \(x^{2}+y^{2}-8x + 6y+25 - 25=0\), which gives \(x^{2}+y^{2}-8x + 6y=0\).

Answer:

The standard form of the equation of the circle is \((x - 4)^2+(y + 3)^2=25\).
The general form of the equation of the circle is \(x^{2}+y^{2}-8x + 6y=0\).