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The center of a circle is $2 a, a − 7$ . If the circle passes through the point (11,-9) and has diameter $102$ . If the circle passes through the point (11,-9) and has diameterunits, then the values of a is and .

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Detailed Solution

The center of a circle is

2 a, a − 7

. If the circle passes through the point (11,-9) and has diameter

102

units, then the values of a is 3 and 5.
We have been given,
The center of a circle is

2 a, a − 7

.
The circle passes through the point (11,-9)
The diameter =

102

units.
Using the distance formula

D = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

.
We know that, the diameter of the circle is

102

.
Therefore, the radius of the circle is,

1022 = 52

The radius of the circle is the distance between points

(2 a, a − 7)

and

(11, − 9)

.
Substituting

x_{2} = 11, x_{1} = 2 a, y_{2} = - 9, and y_{1} = (a - 7)

, in the formula,

∴ r = \sqrt{{(11 - 2 a)}^{2} + {(- 9 - (a - 7))}^{2}}

\Rightarrow \sqrt{121 - 44 a + 4 a^{2} + a^{2} + 4 a + 4}

\Rightarrow \sqrt{5 a^{2} - 40 a + 125}

Solving further,

\sqrt{5 a^{2} - 40 a + 125} = 5 \sqrt{2}

5 a^{2} - 40 a + 125 = 50

5 a^{2} - 40 a + 75 = 0

a^{2} - 8 a + 15 = 0

a^{2} - 5 a - 3 a + 15 = 0

a (a - 5) - 3 (a - 5) = 0

(a - 5) (a - 3) = 0

∴ a = 3, a = 5

Thus, the value of ‘a’ is 3 and 5.

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