Mathematics, 01.09.2019 02:30 oof5221

Which equation represents a circle with a center at (–3, –5) and a radius of 6 units? (x – 3)2 (y – 5)2 = 6 (x – 3)2 (y – 5)2 = 36 (x 3)2 (y 5)2 = 6 (x 3)2 (y 5)2 = 36

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Answer from: ilovepuppieslol4805

The Equation of a circle with a center(-3,-5) and a radius of 6 units is given by;

$(x+3)^{2} +(y+5)^{2} =36$

Further Explanation:Equation of circle An equation of a circle is calculated when the radius of a circle and the coordinates of its center are known or can be calculated.Equation of a circle can be written in two forms namely; the standard and the general form.Standard formStandard form of Equation of a circle is written as;

$(x-a)^{2} +(y-b)^{2} =r^{2}$

Where (a,b) is the center of the circle and r is the radius of the circle.When an equation is written in this form the center and the radius can easily be identified from the equation, with radius always being positive.General form The general form of an equation of a circle is given by;

$x^{2} +y^{2} +dx+ey+f =0$ ; where d, e and f are constants.

When an equation is written in this form one has to use the completing square method to write it in standard form so as to identify the center and the radius.In this case;Center of the circle (a,b) is (-3,-5)

The radius of the circle, r is 6 units

Therefore;

Substituting the values in the general equation we get;

$(x-(-3))^{2} +(y-(-5)^{2} = 6^{2}$

This gives us:

$(x+3)^{2} +(y+5)^{2} =36$

Hence; A circle with a radius of 6 units and a center of (-3,-5) will have an equation of: $(x+3)^{2} +(y+5)^{2} =36$

Keywords: Circle, equation of the circle, center, radius

Learn more about:Equation of a circle; Standard form of equation of a General form of equation of a circle;

Level: High school

Subject: Mathematics

Topic: Equation of the circle

Answer from: aedmund1225

The equation of a circle with center at (-3,-5) , radius of units is $\fbox{\begin\\\ \math (x+3)^{2}+(y+5)^{2}=36\\\end{minispace}}$

Further explanation:

A circle is defined as a locus of all such points which are equidistant from a particular point called the center.

The standard equation of a circle of radius units and the center at (h,k) is as follows:

$\fbox{\begin\\\ \math (x-h)^{2}+(y-k)^{2}=r^{2}\\\end{minispace}}$

Label the above equation as equation (1).

$(x-h)^{2}+(y-k)^{2}=r^{2}$ (1)

It is given that the center of the circle is (-3,-5) and the radius of the circle is units.

Since, the center of a circle is represented as (h,k) and the radius is represented as . So, the value of is , value of is and the value of is units.

To obtain the equation of the circle substitute for , for and for in equation (1).

$\fbox{\begin\\\ \math (x+3)^{2}+(y+5)^{2}=36\\\end{minispace}}$

Figure 1 (attached in the end) represents the graph of the circle $(x+3)^{2}+(y+5)^{2}=36$ .

Therefore, the equation of a circle with center at (-3,-5) , radius of units is $\fbox{\begin\\\ \math (x+3)^{2}+(y+5)^{2}=36\\\end{minispace}}$ .

Learn more:

1.A problem to complete the square of quadratic function

2.A problem to determine the slope intercept form of a line

3.Inverse function

Answer details

Grade: High school

Subject: Mathematics

Chapter: Circle

Keywords: Circle, equation, locus, points, radius, center, equidistant, equation of circle, geometry, (x+3)2+(y+5)2=36, (-3,-5), radiius 6 units.

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hope this : )

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idk

step-by-step explanation:

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