# find the equation of a circle through (0,5),(3,4), touching the line y+5=0.

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### 2 Answers

The general equation of a circle has three variables, if the center is (a, b) and r is the radius: (x - a)^2 + (y - b)^2 = r^2.

You have provided two points through which the circle passes (0,5) and (3,4) and a tangent line to the circle y + 5 = 0

Now, as the line y + 5 = 0 is a tangent, the y-coordinate of the point that touches the line is -5. The point (0, 5) lies on the other end of the circle. This gives the radius of the circle as 5.

So we have a^2 + (5 - b)^2 = 25

and (3 - a)^2 + (4 - b)^2 = 25

(3 - a)^2 - a^2 + (4 - b)^2 - (5 - b)^2 = 25

=> (3 - a - a)(3 - a + a) + (4 - b - 5 + b)(4 - b + 5 - b) = 0

=> 3(3 - 2a) - 1(9 - 2b) = 0

=> 9 - 6a - 9 + 2b = 0

=> 6a = 2b

=> 3a = b

Substitute in a^2 + (5 - b)^2 = 25

=> a^2 + (5 - 3a)^2 = 25

=> a^2 + 25 + 9a^2 - 30a = 25

=> 10a^2 - 30a = 0

=> 10a(a - 3) = 0

=> a = 0 and a = 3

b = 0 and b = 9

**The equation of the required circle can be x^2 + y^2 = 25 and (x - 3)^2 + (y - 9)^2 = 25**

We'll write the equation of the circle:

(x-h)^2 + (y-k)^2 =r^2

h and k are the coordinates of the center of the circle.

If the points (0,5),(3,4) are located on the circle, then their coordinates verify the equation of the circle.

(0-h)^2 + (5-k)^2 =r^2

(3-h)^2 + (4-k)^2 =r^2

Since the circle is tangent to the line y+5=0 => y=-5, then the radius of the circle is 5.

h^2 + 25 - 10k + k^2 = 25

h^2 - 10k + k^2 = 0

h^2 + k^2 = 10k

9 - 6h + h^2 + 16 - 8k + k^2 = 25

25 - 6h - 8k + 10k = 25

-6h + 2k = 0

-3h + k = 0

k = 3h

h^2 + 9h^2 = 30h

10h^2 = 30h

h^2 = 3h

h^2 - 3h = 0

h(h - 3) = 0

h1 = 0

h2 = 3

For h = 0 => k=0

For h = 3 => k=9

**The equations of the circle are: x^2 + y^2 = 25 and (x-3)^2 + (y-9)^2 = 25.**