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The general form of the equation of a circle is Ax2 + By2 + Cx + Dy + E = 0, where A = B 0. If the circle has a radius of 3 units and the center lies on the y-axis, which set of values of A, B, C, D, and E might correspond to the circle?

Aa=0 b=0 c=2 d=2 e=3
Ba=1 b=1 c=8 d=0 e=9
Ca=1 b=1, C = 0, D = -8, and E = 7
D

A = 1, B = 1, C = 8, D = 8, and E = 3

Verified Answer :

Answer:

option: C

Step-by-step explanation:

The general form of a equation of a circle is given as:

\(Ax^2+By^2+Cx+Dy+E=0\)

; We are given that A=B and the radius of circle is 3 units also the centre lies on y-axis.

now we consider option 3

\(A=1,B=1,C=0,D=-8 ,E=7.\)

on putting the values of A,B,C,D and E in equation of circle we get

\(x^{2}+y^{2}-8y+7=0\\ \\x^2+y^2+16-16-8y+7=0\\\\x^{2}+(y-4)^2-16+7=0\)

\(x^{2}+(y-4)^2-9=0\\ \\x^{2} +(y-4)^2=9\\\\x^{2} +(y-4)^2=3^2\)

Hence, the given properties are satisfied.

Hence, option C is correct.

The value will be " a=1 b=1, C = 0, D = -8, and E = 7".

Equation of circle:

\(x^2+y^2-8y+7=0.\)

\(Ax^2 + By^2+ Cx + Dy+ E= 0 \\\\r=3\)

A=B

A, B, C, D, and E=?

\((x-a)^2 + (y-b)^2 = r^2\) is another variant of the circle equation.

The center coordinates are a and b, and the radius is r.

So the equation is \(x^2+(y-4)^2=9\).

The usual form:

\(So\\ A=1, B=1, C=0, D=-8,\ \ and\ \ E=7.\)

OR

\(x^2 + y^2 - 8y + 7 = 0 \\\\x^2 + y^2 + 16 - 16 - 8y + 7 = 0\\\\ x^2 + (y - 4)^2 - 16+ 7 = 0 \\\\x^2 + (y - 4)^2 - 9 = 0\\\\ x^2 + (y - 4)^2 = 9\\\\ x^2 + (y - 4)^2 = 3^2\)

Therefore, the final answer is "Option C".

Find out more information about the circle equation here:

brainly.com/question/10165274

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