Login

Welcome, Guest. Please login or register.

March 30, 2024, 01:16:32 am

Author Topic: Does anyone know where I went wrong?  (Read 477 times)  Share 

0 Members and 1 Guest are viewing this topic.

Rose34

  • Trendsetter
  • **
  • Posts: 158
  • Respect: +2
Does anyone know where I went wrong?
« on: March 09, 2020, 07:05:15 pm »
0
 Find the equation of the circle with centre (2,−3) which touches the x-axis.

My working out:
(x-h)^2 +(y-k)=r^2
(0-2)^2 + (0--3)^2= r^2
4+9=r^2
13=r^2
(x-2)^2 + (y+3)^2= 13

But apparently the correct answer is
 (x−2)^2 + (y + 3)^2 = 9

Thanks in advance!!

S_R_K

  • MOTM: Feb '21
  • Forum Obsessive
  • ***
  • Posts: 487
  • Respect: +58
Re: Does anyone know where I went wrong?
« Reply #1 on: March 09, 2020, 07:26:37 pm »
0
(0-2)^2 + (0--3)^2= r^2

By substituting in x=0 and y=0, you are assuming that the circle passes through (0, 0), but that is not given in the question.

Notice that, in general, if a circle just *touches* an axis (ie, does not cut through at two points), then the distance from the centre of the circle to that axis-intercept is equal to the radius.

Rose34

  • Trendsetter
  • **
  • Posts: 158
  • Respect: +2
Re: Does anyone know where I went wrong?
« Reply #2 on: March 09, 2020, 07:42:29 pm »
0
By substituting in x=0 and y=0, you are assuming that the circle passes through (0, 0), but that is not given in the question.

Notice that, in general, if a circle just *touches* an axis (ie, does not cut through at two points), then the distance from the centre of the circle to that axis-intercept is equal to the radius.

Thanks for the help I thought when it said that it *touches* it meant (0,0)