Mathematics
06.08.2020 02:21
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What is the radius of a circle whose equation is x2+y? +8x-6y+21=0? 2 units 3 units

What is the radius of a circle whose equation is x2+y? +8x-6y+21=0?
2 units
3 units
4 units
5 units
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ptrlvn01
ptrlvn01
4,5(20 marks)

  2 units

Step-by-step explanation:

Completing the squares for the x- and y-terms, we get ...

  (x^2 + 8x) +(y^2 -6y) = -21

  (x^2 +8x +16) +(y^2 -6y +9) = -21 +16 +9

  (x +4)^2 +(y -3)^2 = 2^2

Compare this to the standard form equation for a circle:

  (x -h)^2 +(y -k)^2 = r^2

and we see that the radius (r) is 2.


What is the radius of a circle whose equation is x2+y? +8x-6y+21=0?  2 units 3 units 4 units 5 units
gdubak
gdubak
5,0(69 marks)
2 units.

Step-by-step explanation:

The given equation is

x^{2}+y^{2}+8x-6y+21=0

To find the radius of the circle, we have to complete the squares for both variables, that would give us the explicit form, through which we can determine the radius.

x^{2}+y^{2}+8x-6y+21=0\\x^{2}+8x+y^{2}-6y=-21\\x^{2}+8x+(\frac{8}{2})^{2}+y^{2}-6y+(\frac{6}{2})^{2}=-21+(\frac{8}{2})^{2}+(\frac{6}{2})^{2}\\x^{2}+8x+16+y^{2}-6y+9=-21+16+9\\ (x+4)^{2}+(y-3)^{2}=4\\ (x+4)^{2}+(y-3)^{2}=2^{2}

So, according to this result the radius is 2, because the explicit expression of a circle states:

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

Where the right part of the equation is the square power of the radius.

Therefore, the radius is 2 units.

arnold2619
arnold2619
4,9(73 marks)
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