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# ​ Math help pls

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1 pls help

Feb 16, 2019

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$$(x-h)^2+(y-k)^2=r^2$$ is the center-radius form for a circle such that $$(h,k)$$ is the coordinates of the center of the circle on the coordinate plane and $$r$$ is the radius.

1) Find the equation of the circle centered at (1,-4) with radius 3

Since $$(h,k)$$  of $$(x-h)^2+(y-k)^2=r^2$$ represents the coordinates of the center and $$(1,-4)$$ is already known to be the center, then we know that $$h=1\text{ and }k=-4$$$$r$$ is the radius, so r=3.

Substitute these values into the center-radius form of a circle to generate the equation of this particular circle.

$$h=1,k=-4,r=3\\ (x-h)^2+(y-k)^2=r^2\\ (x-1)^2+(y-(-4))^2=3^2\\ (x-1)^2+(y+4)^2=9$$

2) I encourage you to use the information I gave you above to try and work out the second problem.

3) Find the distance between the points $$(0,8)\text{ and }(2,-5)$$.

We can use the distance formula, $$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$$ . Substitute the points that we want to find the distance from.

 $$d=\sqrt{(8-(-5))^2+(0-2)^2}\\ d=\sqrt{13^2+(-2)^2}\\ d=\sqrt{169+4}\\ d=\sqrt{173}$$ Substitute in the known coordinates and simplify completely. 173 has no perfect square factors, so the radical cannot be simplified any further.

Feb 17, 2019