What is the equation of the circle in general form?x2+y2−6x+10y+22=0x squared plus y squared minus 6 x plus 10 y plus 22 equals 0x2+y2−6x+10y+25=0x squared plus y squared minus 6 x plus 10 y plus 25 equals 0x2+y2+6x−10y+22=0x squared plus y squared plus 6 x minus 10 y plus 22 equals 0x2+y2+6x−10y+25=0x squared plus y squared plus 6 x minus 10 y plus 25 equals 0A circle on a coordinate plane centered at begin ordered pair 3 comma negative 5 end ordered pair. The horizontal x-axis ranges from negative 10 to 10 increments of 1. The vertical y-axis ranges from negative 10 to 8 in increments of 1. Circle touches vertical axis at y equals negative 5. The circle passes through begin ordered pair 0 comma negative 5 end ordered pair, begin ordered pair 3 comma negative 2 end ordered pair, begin ordered pair 6 comma negative 5 end ordered pair, and begin ordered pair 3 comma negative 8 end ordered pair.

What is the equation of the circle in general formx2y26x10y220x squared plus y squared minus 6 x plus 10 y plus 22 equals 0x2y26x10y250x squared plus y squared class=

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explanation

The equation of a circle is given by

[tex](x-a)^2+(y-b)^2=r^2[/tex]

Where a and b are the centres of the circle

In our case we have

[tex]\begin{gathered} a=3 \\ b=-5 \end{gathered}[/tex]

R is the radius of the circle

We have the diameter as 6

Then the radius will be 3

Also, the value of r =3

So we can obtain the equation

[tex](x-3)^3+(y-(-5))^2=3^2[/tex]

[tex]x^2-6x+9+y^2+10y+25=9[/tex]

Simplifying further

[tex]x^2+y^2-6x+10y+25=0[/tex]

The answer is

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