Question

22. 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✓2 units.​

Answer 1

Given ,

The centre of circle is (2a , a - 7) and passes through the point (11,-9)

Diameter of circle = 10√2 units

As we know that ,

Diameter = 2 × radius

And

$\boxed{ \tt{ {(r)}^{2} = {(x - h)}^{2} + {(y - k)}^{2} }}$

Where ,

(x , y) = Any point on circle

(h , k) = Centre of circle

Thus ,

$\sf \implies {(5 \sqrt{2} )}^{2} = {(11 - 2a)}^{2} + { ( - 9 - \{a - 7 \})}^{2}$

$\sf \implies 50 = 121 + 4 {a}^{2} - 44a + { (- a - 2)}^{2}$

$\sf \implies 50 = 121 + 4 {a}^{2} - 44a + { ( - a)}^{2} + 4 - 4( - a)$

$\sf \implies 71 + 5 {a}^{2} - 40a + 4 = 0$

$\sf \implies 5 {a}^{2} - 40a + 75 = 0$

$\sf \implies {a}^{2} - 8a + 15 = 0$

$\sf \implies {a}^{2} - 5a - 3a + 15 = 0$

$\sf \implies a(a - 5) - 3(a - 5) = 0$

$\sf \implies a - 3 = 0 \: \: or \: \: a - 5 = 0$

$\sf \implies a = 3 \: \: or \: \: a = 5$

Hence , the value of a can be 3 or 5

Answer 2

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