Question

Find the equation of a circle of radius 5 units and the centre lies on the X axis and
which passes through the point (2, 3).
​

Answer 1

Answer:

Given : one root of the equation 2x² + Px - 5 = 0 is - 5 and the quadratic equation p(x² + x) + k = 0 has equal roots.

To find : The value of k

solution : one root of the equation 2x² + px - 5 = 0 is -5

so, 2(-5)² + p(-5) - 5 = 0

⇒2 × 25 - 5p - 5 = 0

⇒45 - 5p = 0

⇒p = 9

now the quadratic equation p(x² + x) + k = 0

⇒9(x² + x) + k = 0

⇒9x² + 9x + k = 0

Discriminant = (9)² - 4(9)(k) = 0 [ for equal roots ]

⇒81 - 36k = 0

⇒k = 81/36 = 9/4

Therefore the value of k is 9/4

Answer 2

Given:-

• The equation of a circle of radius 5 units and the centre lies on the x axis.
• The point passes through (2, 3)

To find:-

• Find the equation of the circle..?

Solutions:-

• Let the equation of the required circle be (x - h)² + (y - k)² = r²

Since the radius of the circles is 5 and it's centre lies on the x - axis.

• k = 0 and r = 5

Now, the equation of the circle become (x - h)² + y = 25

Given that, the circles passes through point (2, 3)

Therefore,

=> (2 - h)² + (3)² = 25

=> (2 - h)² + 9 = 25

=> (2 - h)² = 25 - 9

=> (2 - h)² = 16

=> 2 - h = √16

=> 2 - h = +,- 4

So,

=> 2 - h = 4 or h = -2

=> 2 - h = -4 or h = 6

Now, h = -2, the equation of the circle.

=> (x + 2)² + y² = 25

=> x² + 4x + 4 + y² = 25

=> x² + y² + 4x - 21 = 0

h = 6, the equation of the circle.

=> (x - 6)² + y² = 25

=> x² - 12x + 36 + y² = 25

=> x² + y² - 12x + 11 = 0

Hence, the equation of the required circle is x² + y² + 4x - 21 = 0 and x² + y² - 12x + 11 = 0