Question

Choose the correct alternative:

Equation of a circle which passes through (3, 6) and touches the axes is ______.

Options

• x² + y² + 6x + 6y + 3 = 0

• x² + y² − 6x − 6y − 9 = 0

• x² + y² − 6x − 6y + 9 = 0

• x² + y² − 6x + 6y − 3 = 0

Answer 1

Equation of a circle which passes through (3, 6) and touches the axes is x² + y² − 6x − 6y + 9 = 0

Explanation:

given:

The circle passes through the point (3,6)
The circle touches both axes (x-axis and y-axis)

If a circle touches both x-axis and y-axis, its center lies on the line x = r, y = r (i.e., center is at (r,r)), and radius = r.

(x − r)² + (y − r)² = r²

Plug in point (3, 6)

(3 − r)² + (6 − r)² = r²

(9 − 6r + r²) + (36 − 12r + r²) = r²

(9 + 36) − (6r + 12r) + (r² + r²) = r² ⇒ 45 − 18r + 2r² = r²

2r² − 18r + 45 = r² ⇒ r² − 18r + 45 = 0

r² − 18r + 45 = 0

`r = (18 +- sqrt((-18)^2 - 4(1)(45)))/(2(1))`

`= (18+-sqrt(324-180))/2`

`= (18 +- sqrt144)/2`

`= (18 +-1)/1`

So, r = 15 or r = 3

Write equation using valid radius

Try r = 3

(x − 3)² + (y − 3)² = 9

x² − 6x + 9 + y² − 6y + 9 = 9 ⇒ x² + y² − 6x − 6y + 9 = 0

= x² + y² − 6x − 6y + 9 = 0