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  • 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 sqroot 2 units.

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\sqrt{2}  units.

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Solution
 
distance\: formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}
Given points area A(2a, a –7) and B(11, –9)
Diameter= 10\sqrt{2}
Radius= \frac{10\sqrt{2}}{2}= 5\sqrt{2}
Distance= 5\sqrt{2}
(x1, y1) = (2a, a - 7)                  (x2, y2) = (11, -9)
= \sqrt{\left ( 11-2a \right )^{2}+\left ( -9-a+7 \right )^{2}}= \left ( 5\sqrt{2} \right )

Squaring both sides
 \left ( 11-2a \right )^{2}+\left ( -2-a \right )^{2}= \left ( 5\sqrt{2} \right )^{2}
121 + 4a2 – 44a + 4 + a2 + 4a = 50
502 – 40a + 125 – 50 = 0
5a2 – 40a + 75 = 0
Dividing by 5 we get
a2 – 8a + 15 = 0
a2 – 5a - 3a + 15 = 0
a(a – 5) – 3(a – 5) = 0
(a – 5) (a – 3) = 0
a = 5, 3

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