Question

2. Find the centre and radius of the circle.
(1) (x-5)^2+(y-3)^2=20​

Answer 1

Question:

Find the center and the radius of the circle is its equation is given as;

(x -5)^2 + (y-3)^2 = 20 .

Answer:

• Center is (5,3).

• Radius is 2√5 units.

Note:

The standard equation of a circle with radius"r" and center (h,k), is given by;

(x - h)^2 + (y - k)^2 = r^2

Solution:

Given:

The given equation of the circle is ;

=> (x -5)^2 + (y-3)^2 = 20

=> (x -5)^2 + (y-3)^2 = (√20)^2

=> (x -5)^2 + (y-3)^2 = (2√5)^2

Now,

Comparing the equation of given circle {ie; (x -5)^2 + (y-3)^2 = 20} with the standard equation of the circle

{ie; (x - h)^2 + (y - k)^2 = r^2} , we get;

h = 5

k = 3

r = 2√5

Hence,

Hence,Hence,The center of the given circle is (5,3) and the radius of the given circle is 2√5 units.

Answer 2

Question →

Q. find the centre and radius of the circle.

$\star \: \bf{{(x - 5)}^{2} + {(y - 3)}^{2} = 20}$

Answer→

• Centre is (5,3)
• Radius is √(20) or 2√5 cm.

Step - by - step explanation→

Used property→

Here ,we used only the property of standard equation of a circle ,whose radius is "r" cm and whose centre is ( h, k) .

$\star \: \bf{ {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2} }$

Solution →

Let ,the centre of the given Circle is (h,k) and radius is r cm.

The given equation of circle is →

$\implies \: \bf{ {(x - 5)}^{2} + {(y - 3)}^{2} = 20}$

Comparing this equation and standard equation of circle,

After comparing ,

we get,

$\implies \: \bf{h \: = 5 \: \: and \: k = 3}$

The centre of the given Circle is (5,3).

And,

$\implies \: \bf{ {r}^{2} = 20} \\ \\ \bf{ taking \: \sqrt \: on \: both \: sides } \\ \\ \implies \: \bf{ \sqrt{ {r}^{2} } = \sqrt{20} } \\ \\ \implies \: \boxed{\bf{ r = 2\sqrt{5} }}$

Therefore,

The centre of the given Circle is (5,3).

And, the radius of the given Circle is √(20) or 2√5 cm.