I need help on a problem, if anyone can help. the problem is: Write an equation of the circle that has center (2, 4) and passes through the point (6,7). any help would be nice!
2006-09-20 13:22:34 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
can help without giving it all away
eqtn of a circle with center at (0,0) is x^2 + y^2 = r^2
where r is the radius.
if you are working with eqtns of circles you must have worked with translation of points/objects in cartesian space.
so i'll give you one more hint (x-h)^2 + (y-k)^2 = r^2
in order to learn please do the rest yourself :)
2006-09-20 13:32:11 · answer #1 · answered by xkey 3 · 1⤊ 0⤋
The equation of a circle is a quadratic in x and y.
The circle consists of points that are the same distance (the radius) from the center.
Since the circle goes through (6,7), its radius must be the distance from (2,4) to (6,7). Using the Pythagorean theorem, the distance is the square root of ((6 - 2)^2 + (7 - 4)^2). You can calculate that number, right?
Then the equation will simply express the idea that the distance from (2,4), the center of the circle, to (x,y), a point on the circle, is equal to the value you calculated in the preceding paragraph. So the square root of ((x - 2)^2 + (y - 4)^2) equals the square root you calculated above.
Or, you could also say that (x - 2)^2 + (y - 4)^2 is equal to the above number BEFORE you take the square root. And that is the way you should express it when you turn in your homework.
You need to understand this in order to get through the course, so I haven't just given you the answer without any effort on your part.
2006-09-20 20:33:56 · answer #2 · answered by actuator 5 · 0⤊ 0⤋
When you know the center, it is not hard to find the radius.
(x2 -x1) squared plus ( y2 - y1) squared = radius ^ squared, in other words; a^2+b^2=c^2, look familiar? You probably should look at my profile, I had a question that involved finding an equation of a circle and not knowing the center or the radius, just three coordinates because if you have to solve for the center, you have to have three to lock the circle into one place on the xy grid, otherwise if you only have two, the circle can be anywhere between the two coordinates, it takes knowing that third one to form an equation.
2006-09-21 10:05:17 · answer #3 · answered by honest abe 4 · 0⤊ 0⤋
Since the circle is centered at (2,4) it will look: (x-2)^2+(y-4)^2=r^2
to determine radius, determine the distance between (2,4) and (6,7). = sqrt(25)=5.
Therefore, answer equals (x-2)^2+(y-4)^2=25
2006-09-20 20:30:15 · answer #4 · answered by bruinfan 7 · 0⤊ 0⤋
center (2,4)
passes through point (6,7)
so the radius is: sqr( (6-2)^2 + (7-4)^2 )
= sqr ( 16+ 9 ) =sqr(25)=5
(x-2)^2+(y-4)^2 = 25
2006-09-20 20:26:54 · answer #5 · answered by locuaz 7 · 0⤊ 0⤋