A Circle with Center at (-1,3) passes through the point (-1,4). Find the equation?

Question

The answer is (x+1)^2+(y-3)^2=25. How did they get the radius 25?

The answer was found at the back of my text.

Answer 1

this relates to the pythagoras theorem.

based on the equation, the radius should be 5 units.

An equation of a circle is in the form

(x-a)² + (y-b)² = r²

where the centre of the circle has coordinates (a,b) and has radius r.



However, there seems to be an error in the question.. perhaps a typo error?

Answer 2

The standard form for the equation of a circle with center (h,k) and radius r is:

(x - h)² + (y - k)² = r²

(h,k) = (-1,3)

So all that remains is to find r. It is the distance between the center of the circle and a point on the circumference. This is the distance between (-1,3) and (-1,4). Since the x value is unchanged, it is just the difference between the y values:

r = 4 - 3 = 1

So the equation of the circle is:

(x + 1)² + (y - 3)² = 1²
(x + 1)² + (y - 3)² = 1

The "answer" above is incorrect. It looks like someone took the difference between -1 and 4 as the radius by mistake instead.

Answer 3

The circle would be something like x^2 + y^2 + Ax + by technique of + C = 0 (you do not want coefficients for x^2 and y^2 because of the fact they may be the comparable in a circle, wherein case you will possibly purely divide by using by technique of them). Plug x=2, y=2 in to get 4 + 4 + 2A + 2B + C = 0, or 2A + 2B + C = -8. Do the comparable for the different 2 factors and you will possibly have a 3x3 device of equations. sparkling up that besides the undeniable fact which you be attentive to how!

Answer 4

The circle formula is (x - h)^2 + (y - k)^2 = r^2

(h,k) = center
(x,y) is the point it goes through

(-1 - (-1))^2 + (4 - 3)^2 = r^2
(0)^2 + 1^2 = r^2
0 + 1 = r^2
r = 1

(x + 1)^2 + (y - 3)^2 = 1

Answer 5

the radius should be 1
or the 25 should be 1... i don't think it's a radius

Answer 6

What did you just say? and why did you leave the restroomdorr open?