The line is 2x-Y+1=0 is tangent to the circle at thf point (2,5).and the centre of the circle is on the line x+y=9 find the equastion of the circle.
To find the equation of the circle, we need to determine the coordinates of its center and its radius.
First, let's find the center of the circle. We know that the center lies on the line x+y=9. This line can be rewritten as y = 9 - x.
Let's substitute this value of y into the equation of the circle's tangent line: 2x - y + 1 = 0.
Substituting y with (9 - x) in the equation, we get: 2x - (9 - x) + 1 = 0.
Simplifying the equation, we have: 2x - 9 + x + 1 = 0.
Combining the x terms, we get: 3x - 8 = 0.
Now, solve for x: 3x = 8.
Divide both sides by 3: x = 8/3.
Substituting this value of x back into the equation y = 9 - x, we can find y: y = 9 - (8/3).
Simplifying, we have: y = (27 - 8)/3 = 19/3.
Therefore, the center of the circle is located at the coordinates (8/3, 19/3).
Next, we need to find the radius of the circle. Since the tangent line intersects the circle at the point (2, 5), the distance between the center (8/3, 19/3) and (2, 5) is equal to the radius.
Using the distance formula, we can calculate the radius:
r = √((x2 - x1)^2 + (y2 - y1)^2)
r = √((2 - 8/3)^2 + (5 - 19/3)^2)
r = √((6/3 - 24/3)^2 + (15/3 - 57/3)^2)
r = √((-18/3)^2 + (-42/3)^2)
r = √((6)^2 + (-14)^2)
r = √(36 + 196)
r = √(232)
r ≈ 15.26 (rounded to two decimal places)
Now that we have the center and the radius, we can write the equation of the circle in the standard form (x - h)^2 + (y - k)^2 = r^2.
Using the coordinates of the center (8/3, 19/3) and the radius 15.26, the equation of the circle is:
(x - 8/3)^2 + (y - 19/3)^2 = (15.26)^2