What is the equation of a circle with center (2, -1) and which touches the point (10, 12)

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation of a circle with center (2, -1) and which touches the point (10, 12) has to be determined. If the circle touches the point the radius of the circle is equal to the distance between the points (2, -1) and (10, 12).

The distance between two points (x1, y1) and (x2, y2) is given by `sqrt((x1 - x2)^2 + (y1 - y2)^2)` .

The distance between (2, -1) and (10, 12) is equal to `sqrt((2 - 10)^2 + (-1-12)^2)`

=> `sqrt((-8)^2 + (-13)^2)`

=> `sqrt(64 + 169)`

=> `sqrt 233`

The equation of a circle with center (h, k) and with radius r is (x - h)^2 + (y - k)^2 = r^2

Using the information about the circle, its equation is (x - 2)^2 + (y + 1)^2 = 233

The required equation of the circle is ion (x - 2)^2 + (y + 1)^2 = 233

We’ve answered 318,957 questions. We can answer yours, too.

Ask a question