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3 years ago

What is the radius of a circle whose equation is X^2 plus Y^2 -10X +6 X +18=0?

Mathematics

1 answer:

frosja888 [35]3 years ago

3 0

ANSWER

The radius is 4

EXPLANATION

The given equation is:

${x}^{2} + {y}^{2} - 10y + 6x + 18 = 0$

We complete the square to get the expression in standard form:

${x}^{2} + 6x + {y}^{2} - 10y + 18 = 0$

${x}^{2} + 6x + 9 + {y}^{2} - 10y + 25 = - 18 + 9 + 25$

We factor using perfect squares to get:

${(x + 3)}^{2} + {(y - 5)}^{2} = 16$

This implies that,

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

Comparing to

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

The radius is r=4

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Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by: