SOLUTION: find the equation of circle whose center lies on the line x=2y;passes through (2,1)and the length of the tangent from origin is radical 7(seven) ; please i am so tierd with this qu

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the equation of circle whose center lies on the line x=2y;passes through (2,1)and the length of the tangent from origin is radical 7(seven) ; please i am so tierd with this qu      Log On


   

Question 697785: find the equation of circle whose center lies on the line x=2y;passes through (2,1)and the length of the tangent from origin is radical 7(seven) ; please i am so tierd with this question help me
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!




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+%28sqrt%287%29%29%5E2 

(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 = 6%2F5

Using (2), h = 2·6%2F5
           h = 12%2F5

So the center C(h,k) = C(12%2F5,6%2F5)

We just need to find r²

Using (4)

          h²+k² = r²+7
         %2812%2F5%29%5E2+%286%2F5%29%5E2 = r²+7
         144%2F25+36%2F25 = r²+7
         180%2F25 = r²+7
         36%2F5 = r²+7
         36%2F5-7 = r²
         36%2F5-35%2F5 = r²
         1%2F5 = r²

So we can write the equation of the circle as

        (x-h)² + (y-k)² = r²
        {x-12%2F5)² + (y-6%2F5)² = 1%2F5

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

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!


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+%28sqrt%287%29%29%5E2 

(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 = 6%2F5

Using (2), h = 2·6%2F5
           h = 12%2F5

So the center C(h,k) = C(12%2F5,6%2F5)

We just need to find r²

Using (4)

          h²+k² = r²+7
         %2812%2F5%29%5E2+%286%2F5%29%5E2 = r²+7
         144%2F25+36%2F25 = r²+7
         180%2F25 = r²+7
         36%2F5 = r²+7
         36%2F5-7 = r²
         36%2F5-35%2F5 = r²
         1%2F5 = r²

So we can write the equation of the circle as

        (x-h)² + (y-k)² = r²
        {x-12%2F5)² + (y-6%2F5)² = 1%2F5

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