Question

The centre of a circle is (2a – 1, 7) and it passes through the point (-3, -1). If the diameter of the circle is 20 units, then find the value of a

Answer 1

GIVEN :-

• Centre of Circle = (2a - 1 , 7)
• Point on the Circle = (-3 , -1)
• Diameter of Circle = 20 units.

TO FIND :-

• The Value of a.

SOLUTION :-

From the question we have, The centre of the circle (2a - 1 , 7) passes through the point (-3 , -1). Hence The distance between the centre and the point will be equal to the radius.

➔ Diameter (AB) = 20 units.

➔ Radius (OA) = 20/2 = 10 units.

Now by using Distance Formula,

$\mapsto \sf \: OA = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Here ,

➔ x2 = 2a - 1

➔ x1 = -3

➔ y2 = 7

➔ y1 = -1.

Now substitute the above values ,

$\mapsto \sf 10 =\sqrt{ \{2a - 1 - ( - 3) ^{2} \} + \{7 - ( - 1) \}^{2} } \\ \\ \mapsto \sf 10 = \sqrt{(2a - 1 + 3) ^{2} + (7 + 1) ^{2} } \\ \\ \mapsto \sf 10 = \sqrt{(2a + 2) ^{2} + (8) ^{2} } \\ \\ \mapsto \sf 10 = \sqrt{4a ^{2} + 4 + 8a + 64 } \\ \\ \mapsto \sf 10 ^{2} = 4a ^{2} + 4 + 8a + 64 \\ \\ \mapsto \sf 100 = 4a ^{2} + 4 + 8a + 64 \\ \\ \mapsto \sf 4a ^{2} + 4 + 8a + 64 - 100 = 0 \\ \\ \mapsto \sf 4a ^{2} + 4 + 8a - 36= 0 \\ \\ \mapsto \sf 4a ^{2} + 8a \: - 32 = 0$

By splitting the middle term ,

$\mapsto \sf 4a ^{2} + 16a - 8a - 32 = 0 \\ \\ \mapsto \sf \: 4a(a + 4) - 8(a + 4) = 0 \\ \\ \mapsto \sf(4a - 8)(a + 4) = 0 \\ \\ \mapsto \sf \: 4a - 8 = 0 \: , \: a + 4 = 0 \\ \\ \mapsto \sf \: 4a = 8 \: , \: a = - 4 \\ \\ \mapsto \underline { \boxed{ \red{\sf \: a = 2\: , \: a = - 4}}}$

❑ Hence a = 2 , -4

Answer 2

Given ,

The centre of the circle is(2a - 1, 7)

The circle passes through the point (-3 ,-1)

The diameter of the circle is 20 units

So ,

Radius (r) = 20/2 = 10 units

Now , the standard equation of circle is given by

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

Where ,

• r = radius
• (h , k) = center of circle
• (x , y) = any point on circle

Thus ,

$\tt \implies {(10)}^{2} = {( - 3 - \{2a - 1 \})}^{2} + {( - 1 - 7)}^{2}$

$\tt \implies 100 = {(2a - 2)}^{2} + {( - 8)}^{2}$

$\tt \implies 100 = 4 {(a)}^{2} + 4 - 8a + 64$

$\tt \implies 32 = 4 {(a)}^{2} - 8a$

$\tt \implies 4 {(a)}^{2} - 8a - 32 = 0$

$\tt \implies {(a)}^{2} - 2a - 8 = 0$

$\tt \implies {(a)}^{2} - 4a +2a - 8 = 0$

$\tt \implies a(a - 4) + 2(a - 4) = 0$

$\tt \implies a = -2 \: \: or \: \: a = 4$

The value of a will be -2 or 4