13. Given that  {x}^{2} + {y}^{2} + 10y + 16 = 0 and  {(x - 3)}^{2} + {y}^{2} = 1 are two circles on

Question

13. Given that
 {x}^{2} + {y}^{2} + 10y + 16 = 0
and
 {(x - 3)}^{2} + {y}^{2} = 1 are two circles on the same plane. Find:
a) the coordinates of the center and the radius for each circle.
b) the equation of the straight line joining the center of both circles.​

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bonexptip 2 months 2021-07-26T01:05:07+00:00 1 Answers 2 views 0

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    2021-07-26T01:06:54+00:00

    step by step explanation:

    \mathfrak{x}^{2}+{y}^{2}+16=0

    =[x2+16=0x26]

    =[2x{y}^2{16}~0]

    =[4×{y}^0{16}]

    =[32x{y}^x]

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