Question

The centre of a circle is (2a, a -7). Find the values of a, if the circle passes through the point (11, -9) and has diameter 10 units.

Answer 1

Answer:

Step-by-step explanation:

Let the centre of circle be O(2a, a-7) and the point on the circumference be A(11, -9).

Therefore, x= 2a, y= a-7

                  x'= 11, y'= -9

Since, diameter= 10 units.

Thus, radius= 10/2 = 5 units.

We know that,

A straight line drawn from the centre to any point on the circumference is the radius of the circle.

Thus, OA is a radius.

Also,

Distance between two points= $\sqrt{(x-x')^{2} + (y-y')^{2} }$

   =>               OA                =$\sqrt{(2a-11)^{2}+(a+2)^{2} }$  [since, y-y'= (a-7)-(-9)=                                            

                                                                                          a-7+9= a+2]

      =>               5                   =$\sqrt{(2a)^{2} -2*2a*11 + 11^{2} + a^{2} +2*a*2 + 2^{2} }$

By squaring on both sides, we get,

       =>             $5^{2}$                     =$(\sqrt{(2a)^{2} -2*2a*11 + 11^{2} + a^{2} +2*a*2 + 2^{2}})^{2}$

      =>             25                     = $4a^{2} -44a + 121 +a^{2} + 4a + 4$

     =>             25                      = $5a^{2} - 40a +125$

     =>             0                         = 5$a^{2}$ - 40a +125 - 25

     =>        $5a^{2}$ - 40a + 100        = 0

     =>       5($a^{2}$ - 8a + 20)          = 0

    =>         $a^{2}$ - 8a + 20              = 0

    co-efficient of $a^{2}$= k = 1

    co-efficient of a= l = -8

                                m= 20

Therefore,

by quadratic formula,

         a =   $\frac{-l-\sqrt{l^{2} - 4(k)(m)}}{2k}$ , $\frac{-l+\sqrt{l^{2} - 4(k)(m)}}{2k}$

  =>       a =  $\frac{-(-8)-\sqrt{(-8)^{2} - 4(1)(20)}}{2(1)}$ , $\frac{-(-8)+\sqrt{(-8)^{2} - 4(1)(20)}}{2(1)}$

  =>      a = $\frac{8-\sqrt{64 - 80}}{2}$ , $\frac{8+\sqrt{64 - 80}}{2}$

  =>      a = $\frac{8-\sqrt{-16}}{2}$ , $\frac{8+\sqrt{-16}}{2}$

  =>      a = $\frac{8-\sqrt{(-1)(16)}}{2}$ , $\frac{8+\sqrt{(-1)(16)}}{2}$

 =>       a = $\frac{8-4\sqrt{(-1)}}{2}$ , $\frac{8+4\sqrt{(-1)}}{2}$

 =>       a = $\frac{2(4-2\sqrt{(-1)})}{2}$ , $\frac{2(4+2\sqrt{(-1)})}{2}$

=>        a = $4-2\sqrt{(-1)}$ , $4+2\sqrt{(-1)}$

Therefore, a= $4-2\sqrt{(-1)}$ , $4+2\sqrt{(-1)}$