Does the point (3,-4) lie on a circle whose center is at (-2,3) and passes through (5,8)?

Question

Does the point (3,-4) lie on a circle whose center is at (-2,3) and passes through (5,8)?

what does "passes through means"?....

what does the question means?

Answer 1

You would figure this out by finding the distance from the center of the circle to each point. If each distance is the same, then the point lies on the circle.
To find the distance from center to first point:
sqrrt( [-2-3]^2 + (3- -4)^2 ) = sqrrt (25 + 49) = sqrrt (74)

To find the distance from center to point on circle:
sqrrt( [-2-5]^2 + (3- 8)^2 ) = sqrrt (49 + 25) = sqrrt (74)

So, since both points are the same distance from the center of the circle, then they are both on the circle.

Answer 2

The Equation of a circle with center at (-2,3) is :
(x+2)^2 + (y-3)^2 = r^2, where r is the radius of the circle. If the circle passes through the point (5,8) then the distance from the center to (5,8) is a radius of the circle.

Its length is r^2 = (5-(-2))^2 + (8-3)^2 = 49+25 =74

So the equation of the circle is (x+2)^2 +y-3)^2 = 74

Now put 3 for x and -4 for y and see what you get
(3+2)^2 + (-4-3)^2 = 74
25+49 = 74
74=74
So YES, the point (3,-4) lies on the circle which means the circle passes through it.

Answer 3

Let the center be (x0, y0) and the radius be r

Equation of a circle:
(x - x0)^2 + (y - y0)^2 = r^2

(x0,y0) = (-2,3)
r = distance from the center to a point on the circle (that's what they mean by passes through - on the circle)
= sqrt[(8 - y0)^2 + (5 - (x0))^2] (this is just the distance formula)

(x + 2)^2 + (y - 3)^2 = [sqrt[(8-3)^2 + (5+2)^2]]^2
(x + 2)^2 + (y - 3)^2 = [sqrt[25 + 49]]^2
(x + 2)^2 + (y - 3)^2 = [sqrt[74]]^2
(x + 2)^2 + (y - 3)^2 = 74

Now plug in the point (3,-4). If the right hand side and left hand side of the equation are equal, the point does lie on the circle.

Answer 4

If two points lie on a circle centred at (-2,3), then their distances from that point must be the same, as all points on a circle are equidistant from the centre.

The circle passes through a point if that point lies on the circumference of the circle.

The distance of (5,8) from (-2,3) is:
sqrt( (8 - 3)^2 + (5 + 2)^2 )
= sqrt(25 + 49)
= sqrt(74)

The distance of (3,-4) from (-2,3) is:
sqrt( (-4 - 3)^2 + (3 + 2)^2)
= sqrt(49 + 25)
= sqrt(74).

The point does therefore lie on the circle.

Answer 5

I would hope that the term "pass through" would be fairly obvious; if you aim a BB gun at a hole in a wall, the shot will either pass through the hole or miss it and hit the wall. In this problem, the best way to proceed is to do a coordinate transformation to move the circle center to the origin. This puts the test point at (5, -7), the pass-through point at (7, 5), and since the distance from the origin to each of these is obviously the same (Pythagoras, you know), the answer to the problem is Yes.

Answer 6

passes through means that one of the points on the circle is (5,8), so you find the distance from the center to the point (5,8) then find the distance from the center to the point (3,-4) if they are equal, the point is on the circle

In case you don't have it, the formula for finding the distance is sqrt( (x1-x2)^2 + (y1-y2)^2))

good luck

Answer 7

"Passes through" simply means that (5,8) is a point on the circle.

To find the answer, use the distance formula...

d= square root of [(x2-x1)^2 + (y2-y1)^2]

d=sr (-2-3)^2 + (3--4)^2
d=sr (-5)^2 + (7)^2
d=sr 25+49
d=sr 74

repeat for other point

d=sr (5--2)^2 + (8-3)^2
d=sr 7^2 + 5^2
d=sr 49+25
d=sr 74

The distances (which in this case are the radii) match, so the answer is yes.

Answer 8

From your question, we determine the equation of the circle is
(x+2)^2+(y-3)^2=(5+2)^2+(8-3)^2

let
f(x,y)=(x+2)^2+(y-3)^2-(5+2)^2+(8-3)^2

Now given point(-2,3) will be inside the circle if f(2,3)<=0

or
f(2,3)=(2+2)^2+(3-3)^2-(5+2)^2+(8-3)^2
=4^2-7^2+5^2
=16-49+25
=0
So the point is inside the circle and as f(x,y) = 0 so it lies in its permimeter.

Answer 9

Yes

Answer 10

You don't have to go through the math (unless you wnat to). I just simply brought the problem to my cad program and pluged in the coordinates and found that the answer is no.