Question

Find the equation of the circle of radius 5 cm and whose center lies on the y-axis

and which passes through the point P( 5 , 2 )​

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

• Find the equation of the circle of radius 5 cm and whose center lies on the y-axis and which passes through the point P( 5 , 2 ).. ?

Sᴏʟᴜᴛɪᴏɴ :-

If the center lies on y-axis, x coordinate is zero.

Let the centre be (0, a).

Than,

Equation of circle = (x - 0)² + (y - a)² = 5²

And, given That, P(5,2) lies on the circle.

So,

→ (5 - 0)²+ (2 - a)² = 5²

→ 25 + (2 - a)² = 25

→ (2 - a)² = 0

→ (2 - a) = 0

→ a = 2 .

Therefore,

→ Equation of circle at centre (0,2) = x² + (y - 2)² = 5² . (Ans.)

Answer 2

→GIVEN←

A circle with radius 5 cm, whose center lies on the y-axis, and passes through the point P(5,2).

$\underline{\rule{190}2}$

→TO FIND←

The equation of the given circle.

$\underline{\rule{190}2}$

→SOLUTION←

Let the coordinates of the center of the circle be (0,s).

(As the center lies on the y-axis, so the x-coordinate is 0.)

We know the equation of a circle :-

$\boxed{\sf{(x-0)^2+(y-s)^2=r^2}}$

x is 5 and y is 2, as the point P(5,2) through which the circle passes.

Substituting the values :-

$\sf{\implies (5-0)^2+(2-s)^2=(5)^2}\\\\\sf{\implies (5)^2+(2-s)^2=(5)^2}\\\\\sf{\implies (2-s)^2=0}\\\\\sf{\implies 2-s=0}\\\\\boxed{\sf{\implies s=2}}$

∴ Required equation of the circle = (5 - 0)² + (2 - 2)² = (5)².