Question

A circle has a radius of \sqrt{45} 45 ​ square root of, 45, end square root units and is centered at (-2.4,-4.8)(−2.4,−4.8)(, minus, 2, point, 4, comma, minus, 4, point, 8, ). Write the equation of this circle.

Answer 1

Answer:

[TeX]x^2+y^2+4.8x+9.6y-16.2=0[/TeX]

Step-by-step explanation:

The equation of a circle centre(h,k) of radius r is given by:

r²=(x-h)²+(y-k)²

If radius, [TeX]r=\sqrt{45}[/TeX]

Centre, (h,k)=(-2.4,-4.8)

Substitution into the equation of the circle formula

r²=(x-h)²+(y-k)²

[TeX](\sqrt{45})^{2}=(x-(-2.4))^{2}+(y-(-4.8))^{2}[/TeX]

[TeX]45=(x+2.4)^{2}+(y+4.8)^{2}[/TeX]

[TeX]45=x^2+4.8x+5.76+y^2+9.6y+23.04[/TeX]

[TeX]0=x^2+y^2+4.8x+9.6y+5.76+23.04-45[/TeX]

[TeX]x^2+y^2+4.8x+9.6y-16.2=0[/TeX]

The equation of the circle is therefore given as:

[TeX]x^2+y^2+4.8x+9.6y-16.2=0[/TeX]

Answer 2

Answer:

(x + 2.4)^2 + (y + 4.8)^2 = 45

Step-by-step explanation:

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