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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 = 1; (h,k) = (-3, -4)
the standard form of the equation of this circle is
(type your answer in standard form.)
the general form of the equation of this circle is
(simplify your answer. type an equation.)
use the graphing tool to graph the circle.

Explanation:

Step1: Write the standard form of the circle equation

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\). Given \(h=-3\), \(k = - 4\), and \(r = 1\), substitute these values into the formula: \((x-(-3))^2+(y-(-4))^2=1^2\), which simplifies to \((x + 3)^2+(y + 4)^2=1\).

Step2: Expand the standard - form equation to get the general form

Expand \((x + 3)^2+(y + 4)^2=1\).
First, expand \((x + 3)^2=x^{2}+6x + 9\) and \((y + 4)^2=y^{2}+8y+16\).
Then the equation becomes \(x^{2}+6x + 9+y^{2}+8y + 16=1\).
Combine like terms: \(x^{2}+y^{2}+6x + 8y+9 + 16-1=0\).
So the general form is \(x^{2}+y^{2}+6x + 8y+24 = 0\).

Answer:

The standard form of the equation of this circle is \((x + 3)^2+(y + 4)^2=1\).
The general form of the equation of this circle is \(x^{2}+y^{2}+6x + 8y+24 = 0\).