# Writing standard equation of a circle | Mathematics II | High School Math | Khan Academy

– [Voiceover] So we have a circle here and they specified some points for us. This little orangeish, or,
I guess, maroonish-red point right over here is the
center of the circle, and then this blue point
is a point that happens to sit on the circle. And so with that information,
I want you to pause the video and see if you can figure out
the equation for this circle. Alright, let’s work through this together. So let’s first think about
the center of the circle. And the center of the
circle is just going to be the coordinates of that point. So, the x-coordinate is negative one and then the y-coordinate is one. So center is negative one comma one. And now, let’s think about what the radius of the circle is. Well, the radius is going
to be the distance between the center and any point on the circle. So, for example, for example, this distance. The distance of that line. Let’s see I can do it thicker. A thicker version of that. This line, right over there. Something strange about my… Something strange about my pen tool. It’s making that very thin. Let me do it one more time. Okay, that’s better. (laughs) The distance of
that line right over there, that is going to be the radius. So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean Theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So, if we look at our
change in x right over here. Our change in x as we go from the center to this point. So this is our change in x. And then we could say that
this is our change in y. That right over there is our change in y. And so our change in x-squared plus our change in y-squared is going to be our radius squared. That comes straight out of
the Pythagorean Theorem. This is a right triangle. And so we can say that r-squared is going to be equal to our change in x-squared plus our change in y-squared. Plus our change in y-squared. Now, what is our change in x-squared? Or, what is our change in x going to be? Our change in x is going to be equal to, well, when we go from the radius to this point over here, our x goes from negative one to six. So you can view it as our ending x minus our starting x. So negative one minus negative, sorry, six minus negative one is equal to seven. So, let me… So, we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care
about the absolute value in the change of x, and once you square it it all becomes a positive anyway. So our change in x right over here is going to be positive seven. And our change in y, well, we are starting at, we are starting at y is equal to one and we are going to y is
equal to negative four. So it would be negative four minus one which is equal to negative five. And so our change in y is negative five. You can view this distance right over here as the absolute value of our change in y, which of course would be
the absolute value of five. But once you square it, it doesn’t matter. The negative sign goes away. And so, this is going to
simplify to seven squared, change in x-squared, is 49. Change in y-squared,
negative five squared, is 25. So we get r-squared, we get r-squared is equal to 49 plus 25. So what’s 49 plus 25? Let’s see, that’s going
to be 54, or was it 74. r-squared is equal to 74. Did I do that right? Yep, 74. And so now we can write the
equation for the circle. The circle is going to be
all of the points that are, well, in fact, let me right all of the, so if r-squared is equal to 74, r is equal to the square-root of 74. And so the equation of
the circle is going to be all points x comma y that are this far away from the center. And so what are those points going to be? Well, the distance is going to be x minus the x-coordinate of the center. x minus negative one squared. Let me do that in a blue color. Minus negative one squared. Plus y minus, y minus the y-coordinate of the center. y minus one squared. Squared. Is equal, is going to
be equal to r-squared. Is going to be equal to the
know is going to be 74. 74. And then if we want to
simplify it a little bit, you subtract a negative,
this becomes a positive. So it simplifies to x plus one squared plus y minus one squared is equal to 74. Is equal to 74. And, we are all done.

### 28 thoughts on “Writing standard equation of a circle | Mathematics II | High School Math | Khan Academy”

• February 23, 2016 at 7:35 pm

π

• February 27, 2016 at 2:42 am

π

• March 25, 2016 at 3:38 pm

π

• May 24, 2016 at 9:40 pm

π

• June 17, 2016 at 1:37 am

π

• October 16, 2016 at 5:56 am

π

• October 18, 2016 at 5:36 pm

B)

• January 16, 2017 at 6:56 pm

π

• January 30, 2017 at 4:44 am

π

• March 31, 2017 at 11:41 am

π

• April 4, 2017 at 9:43 am

Thank you ππΎπ

• April 23, 2017 at 1:49 am

π

• May 22, 2017 at 5:21 am

y is is squared? π²

• September 1, 2017 at 2:36 pm

π

• October 31, 2017 at 9:34 am

π

• November 3, 2017 at 2:04 pm

π

• December 12, 2017 at 6:52 pm

π

• February 22, 2018 at 5:48 pm

thank you so much, was so worried for my further maths GCSE!!!! This is so simple and makes perfect sense. Keep it up

• May 20, 2018 at 8:56 am

π

• May 23, 2018 at 9:19 pm

π

• August 2, 2018 at 10:08 pm

π

• October 14, 2018 at 5:39 am

now I am confident doing A/s level math

• January 14, 2019 at 7:01 pm

π

• June 29, 2019 at 6:23 am

pls help me in my homework.. the question is: Reduce the equation of the circle to the general form.
given: x+2(y-x^2)=4+2y^2

what do i do first to make it into a general form? thanks in advance!

• August 21, 2019 at 10:14 pm

Shouldn't it be equal to 74^2