Question

The centre of the circle
$(x + 5) {}^{2} + (y - 3) = 36$
is:​

Answer 1

Correct question

The centre of the circle (x+5)² + (y-3)² = 36 is:

Solution

Given:-

• Equation of a circle as

        (x+5)² + (y-3)² = 36

To find:-

• Centre of the circle

 

Knowledge required:-

• Standard equation of a circle

For a circle with centre C having coordinates (h, k), length of the radius of the circle being 'a' and P(x,y) being any point on the circumference of the circle. The central form of the equation for the circle is given by:

       (x - h)² + (y - k)² = a²

(This gives the relation between the coordinates of any point on the circumference and hence is the required equation of the circle with centre (h, k) and radius equal to 'a')

Solution:-

Comparing the given equation for the circle

(x-(-5))² + (y-3)² = 36

with the standard form of the equation of a circle

(x-h)² + (y-k)² = a²

where (h, k) represent the coordinates of the centre of the circle.

we will get,

h = -5   and  k = 3

Therefore,

• the coordinates of the centre of the circle will be (h, k) = (-5, 3).

Answer 2

Answer:

$\huge\mathbb{\fcolorbox{yellow}{gold}{Solution:–}}$

$\rm→[x-(-5)]²+[y-3]²=6²$

Compairing with equation of circle

$\rm→(x - h)²+(y-k)²=r²$

$\rm→Center \: (-5,3) \: r=6$

Ex:–

$\rm→(x - 1)²+(y+2)²=5$

$\rm→Centre(1,-2) \: r= \sqrt{5}$

$\rm→(x+3)²+y²=2$

$\rm → Center (-3,0) \: r= \sqrt{2}$

$\rm→(x - 1)²+(y - 1)²=1$

$\rm→Centre(1,1) \: r=1$