How do you write the equation for a circle having (3,0) and (-2,-4) as ends of a diameter?

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How do you write the equation for a circle having (3,0) and (-2,-4) as ends of a diameter?

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Answer 1 · 2016-08-31T03:04:48

Equation of circle is $x^{2} + y^{2} - x + 4 y - 6 = 0$ $x^2+y^2-x+4y-6=0$

Explanation:

As the end points of diameter are $(3, 0)$ $(3,0)$ and $(- 2, - 4)$ $(-2,-4)$ , the center is their midpoint i.e. $(\frac{3 - 2}{2}, \frac{0 - 4}{2})$ $((3-2)/2,(0-4)/2)$ or $(\frac{1}{2}, - 2)$ $(1/2,-2)$ .

The distance between $(3, 0)$ $(3,0)$ and $(- 2, - 4)$ $(-2,-4)$ will be

$\sqrt{{(3 - (- 2))}^{2} + {(0 - (- 4))}^{2}}$ $sqrt((3-(-2))^2+(0-(-4))^2)$

= $\sqrt{25 + 16}$ $sqrt(25+16)$

= $\sqrt{41}$ $sqrt41$

Hence diameter is $\sqrt{41}$ $sqrt41$ and radius $\frac{\sqrt{41}}{2}$ $sqrt41/2$ .

As the center is $(\frac{1}{2}, - 2)$ $(1/2,-2)$ and radius is $\frac{\sqrt{41}}{2}$ $sqrt41/2$ , the equation of circle is

${(x - \frac{1}{2})}^{2} + {(y + 2)}^{2} = {(\frac{\sqrt{41}}{2})}^{2}$ $(x-1/2)^2+(y+2)^2=(sqrt41/2)^2$ or

$x^{2} - x + \frac{1}{4} + y^{2} + 4 y + 4 = \frac{41}{4}$ $x^2-x+1/4+y^2+4y+4=41/4$ or

$x^{2} + y^{2} - x + 4 y + 4 + \frac{1}{4} - \frac{41}{4} = 0$ $x^2+y^2-x+4y+4+1/4-41/4=0$ or

$x^{2} + y^{2} - x + 4 y + 4 - 10 = 0$ $x^2+y^2-x+4y+4-10=0$ or

$x^{2} + y^{2} - x + 4 y - 6 = 0$ $x^2+y^2-x+4y-6=0$

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How do you write the equation for a circle having (3,0) and (-2,-4) as ends of a diameter?

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