Let the equation be
(1) (x-h)² + (y-k)² = r²
and since it is on the line 2y=x,
(2) 2k = h
(2,1) is on the circle and also on the line 2y=x
So (2,1) satisfies the equation of the circle, so
(2-h)² + (1-k)² = r²
4-4h+h²+1-2k+k² = r²
(3) h²+k²-4h-2k = r²-5
By the distance formula OC² = h²+k²
OCT is a right triangle because a tangent is perpendicular
to the radius drawn to the point of tangency. So using
the Pythagorean theorem:
OC² = TC² + OT²
h²+k² = r²+
(4) h²+k² = r²+7
Subtracting equation (4) from equation (3) gives
-4h-2k = -12 or
(5) 2h+k = 6
Using (2), (5) becomes
2(2k)+k = 6
4k+k = 6
5k = 6
k =
Using (2), h = 2·
h =
So the center C(h,k) = C(,)
We just need to find r²
Using (4)
h²+k² = r²+7
+ = r²+7
+ = r²+7
= r²+7
= r²+7
-7 = r²
- = r²
= r²
So we can write the equation of the circle as
(x-h)² + (y-k)² = r²
{x-)² + (y-)² =
If you like you can square those out, collect terms and clear of
fractions and you'll end up with
5x² + 5y² - 24x - 12y + 35 = 0
Edwin
Let the equation be
(1) (x-h)² + (y-k)² = r²
and since it is on the line 2y=x,
(2) 2k = h
(2,1) is on the circle and also on the line 2y=x
So (2,1) satisfies the equation of the circle, so
(2-h)² + (1-k)² = r²
4-4h+h²+1-2k+k² = r²
(3) h²+k²-4h-2k = r²-5
By the distance formula OC² = h²+k²
OCT is a right triangle because a tangent is perpendicular
to the radius drawn to the point of tangency. So using
the Pythagorean theorem:
OC² = TC² + OT²
h²+k² = r²+
(4) h²+k² = r²+7
Subtracting equation (4) from equation (3) gives
-4h-2k = -12 or
(5) 2h+k = 6
Using (2), (5) becomes
2(2k)+k = 6
4k+k = 6
5k = 6
k =
Using (2), h = 2·
h =
So the center C(h,k) = C(,)
We just need to find r²
Using (4)
h²+k² = r²+7
+ = r²+7
+ = r²+7
= r²+7
= r²+7
-7 = r²
- = r²
= r²
So we can write the equation of the circle as
(x-h)² + (y-k)² = r²
{x-)² + (y-)² =
If you like you can square those out, collect terms and clear of
fractions and you'll end up with
5x² + 5y² - 24x - 12y + 35 = 0
Edwin