Question

Find the equation of a circle with centre (2,2) and passes through the point (4,5).​

Answer 1

To Find :-

• The equation of a circle.

Solution :-

Given,

• The centre of the circle is (h, k) = (2,2)
• The circle passes through point (4,5).

The radius (r) of the circle is the distance between the points (2,2) and (4,5).

$\implies$ r = √[(2-4)² + (2-5)²]

$\implies$ r = √[(-2)² + (-3)²]

$\implies$ r = √[4+9]

$\implies$ r = √13

As we know that,

The equation of the circle is ;

$\red\bigstar$ (x - h)²+ (y - k)²= r²

[ Put the values ]

$\implies$ (x - h)² + (y - k)² = (√13)²

$\implies$ (x - 2)² + (y - 2)² = (√13)²

$\implies$ x² - 4x + 4 + y² - 4y + 4 = 13

$\implies$ x² + y² - 4x - 4y = 5 $\green\bigstar$

Therefore,

The equation of a circle is x² + y² - 4x - 4y = 5.

Answer 2

To Find :-

• The equation of a circle.

Solution :-

Given,

• The centre of the circle is (h, k) = (2,2)

• The circle passes through point (4,5).

• The radius (r) of the circle is the distance between the points (2,2) and (4,5).

$⟹ r = √[(2-4)² + (2-5)²]$

$\implies r = √[(-2)² + (-3)²]$

$\implies r = √[4+9]$

$\implies r = √13$

As we know that,

The equation of the circle is ;

$\red\bigstar (x - h)²+ (y - k)²= r²$

[ Put the values ]

$\implies(x - h)² + (y - k)² = (√13)²</p><p>$

$\implies(x - 2)² + (y - 2)² = (√13)²$

$\implies x² - 4x + 4 + y² - 4y + 4 = 13$

$\implies x² + y² - 4x - 4y = 5 \green\bigstar$

Therefore,

The equation of a circle is x² + y² - 4x - 4y = 5.