How do you write the standard from of the equation of the circle given center is at (-1,2) and the point (3,4) lies on the circle?

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How do you write the standard from of the equation of the circle given center is at (-1,2) and the point (3,4) lies on the circle?

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Answer 1 · 2018-03-09T20:41:29

${(x + 1)}^{2} + {(y - 2)}^{2} = 20$ $(x+1)^2+(y-2)^2=20$

Explanation:

$the standard form of the equation of a circle is$ $"the standard form of the equation of a circle is"$

$∙ x {(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $•color(white)(x)(x-a)^2+(y-b)^2=r^2$

$where (a, b) are the coordinates of the centre and r$ $"where "(a,b)" are the coordinates of the centre and r"$
$the radius$ $"the radius"$

$here (a, b) = (- 1, 2)$ $"here "(a,b)=(-1,2)$

$the radius is the distance from the centre to a point$ $"the radius is the distance from the centre to a point"$
$on the circumference$ $"on the circumference"$

$to calculate r use the distance formula$ $"to calculate r use the "color(blue)"distance formula"$

$∙ x r = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$let (x_{1}, y_{1}) = (- 1, 2) and (x_{2}, y_{2}) = (3, 4)$ $"let "(x_1,y_1)=(-1,2)" and "(x_2,y_2)=(3,4)$

$\Rightarrow r = \sqrt{{(3 + 1)}^{2} + {(4 - 2)}^{2}} = \sqrt{16 + 4} = \sqrt{20}$ $rArrr=sqrt((3+1)^2+(4-2)^2)=sqrt(16+4)=sqrt20$

$\Rightarrow {(x - (- 1))}^{2} + {(y - 2)}^{2} = {(\sqrt{20})}^{2}$ $rArr(x-(-1))^2+(y-2)^2=(sqrt20)^2$

$\Rightarrow {(x + 1)}^{2} + {(y - 2)}^{2} = 20 \leftarrow equation of circle$ $rArr(x+1)^2+(y-2)^2=20larrcolor(red)"equation of circle"$

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How do you write the standard from of the equation of the circle given center is at (-1,2) and the point (3,4) lies on the circle?

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