Question

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

Answer 1

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Answer 2

Answer:

a = 3 or a = 5

Step-by-step explanation:

Since, the equation of a circle,

$(x-h)^2+(y-k)^2 = r^2$

Where,

(h, k) = center of the circle, r = radius,

Here, h = 2a and k = a - 7, diameter = 10√2 ⇒  r = $\frac{10\sqrt{2}}{2}$ = 5√2 unit

So, the equation of the given circle is,

$(x-2a)^2 + (y-a+7)^2 = (5\sqrt{2})^2$

Since, circle passes through the point (11, -9)

i.e. (11, -9) will satisfy the equation of the circle,

$(11-2a)^2 + (-9-a+7)^2 = 50$

$(11-2a)^2 + (-2-a)^2 = 50$

$121 + 4a^2 - 44a + 4 + a^2 + 4a = 50$

$5a^2 - 40a + 125 - 50$

$5a^2 - 40a + 75 = 0$

$a^2 - 8a + 15 =0$

By middle term splitting,

$a^2 - 5a - 3a + 75=0$

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

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

By zero product property,

a - 3 = 0 or a - 5 = 0

a = 3 or a = 5