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**mathxyz****Member**- From: Brooklyn, NY
- Registered: 2024-02-24
- Posts: 1,053

1. The center of a circle is the point (3, 2). If the point ( 2, 10) lies on this circle, find the standard equation for the circle.

2. Find the standard equation of the circle tangent to the x-axis and with center (3, 5).

3. Find the standard equation of the circle tangent to the y-axis and with center (3, 5).

4. Find the standard equation of the circle passing through the origin and with center (3, 5).

5. The points A( 1, 6) and B(3, 2) are the endpoints of a diameter of a circle. Find the y-intercepts of the circle.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 10,610

The standard general equation for a circle is derived like this.

If the centre is at (a,b) and rhe radius is r and (x,y) is the general point on the circumference:

Horizontal distance from the centre to the point is x-a

Vertical distance from the centre to the point is y-b

There's a right angle between the horizontal and vertical lines so Pythagoras can be applied

You can use this for these questions:

1. You're told a and b and you can use the distance from the centre to the given point as the radius.

2. The tangent must be the x axis so at point (3,0) . The radius must be 5. So, once again, we have the radius and the centre.

3. Similar but now its the y axis.

4. Calculate the distance between (3,5) and (0,0). This is the radius.

5. There's more to do here. You can calculate the length of the diameter and hence get the radius. Also the midpoint of the diameter will be the centre. So that's enough to get the equation.

The y intercepts will have x coordinate zero so substitute x=0 in the equation and solve for y (two solutions).

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**mathxyz****Member**- From: Brooklyn, NY
- Registered: 2024-02-24
- Posts: 1,053

Bob wrote:

The standard general equation for a circle is derived like this.

If the centre is at (a,b) and rhe radius is r and (x,y) is the general point on the circumference:

Horizontal distance from the centre to the point is x-a

Vertical distance from the centre to the point is y-bThere's a right angle between the horizontal and vertical lines so Pythagoras can be applied

You can use this for these questions:

1. You're told a and b and you can use the distance from the centre to the given point as the radius.

2. The tangent must be the x axis so at point (3,0) . The radius must be 5. So, once again, we have the radius and the centre.

3. Similar but now its the y axis.

4. Calculate the distance between (3,5) and (0,0). This is the radius.

5. There's more to do here. You can calculate the length of the diameter and hence get the radius. Also the midpoint of the diameter will be the centre. So that's enough to get the equation.

The y intercepts will have x coordinate zero so substitute x=0 in the equation and solve for y (two solutions).

Bob

Bob,

Problems posted in the Help Forum, I need help with. Problems posted here are for members to enjoy math and practice with.

True math lovers will randomly search for questions to answer and explore.

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