Question

The equation of circle passing through (+2, 3) having centre at (2, 0) is​

Answer 1

Answer:

$radius \: \: \: r = \sqrt{ {(2 - 2)}^{2} + {(3 - 0)}^{2} } \\ = \sqrt{ {3}^{2} } \\ = 3$

Formula : if a circle having centre (h,k) and radius = r then the equation of circle is,

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

So, for this question, the equation is,

${(x - 2)}^{2} + {(y - 0)}^{2} = {3}^{2} \\ {x}^{2} - 4x + 4 + {y}^{2} = 9 \\ {x }^{2} + {y}^{2} - 4x - 5 = 0$

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https://brainly.in/question/11755078?utm_source=android&utm_medium=share&utm_campaign=question

Answer 2

Answer:

First we have to find the radius

We know that,

    $r =\sqrt{(x_{1} ^2-x_{2}^2)+(y_{1}^2 -y_{2}^2) }$

∴  $r= \sqrt{(2^2-2^2)+(3^2-0)}$

   $r =\sqrt{3^2}$

   $r=3$

Given,

The center is at (2,0)

Therefore the equation of circle is

    $(x-x_{1} )^2+(y-y_{1} )^2=r^2$

    $(x-2)^2+(y-0)^2=3^2$

    $(x-2)^2+y^2=9$

                   Hope it helps you^_^