Question

Find the equation of the circle whose center and radius are given.

Answer 1

Answer:  The required equation of the circle is $x^2+y^2-14x+6y+9=0.$

Step-by-step explanation:  We are given to find the equation of the circle with center at the point (7, -3) and radius of length 7 units.

We know that

the standard equation of a circle with center at the point (h, k) and radius of length r units is given by

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

For the given circle,

center, (h, k) = (7, -3)  and  radius, r = 7 units.

From equation (i), we get

$(x-7)^2+(y-(-3))^2=7^2\\\\\Rightarrow (x-7)^2+(y+3)^2=49\\\\\Rightarrow x^2-14x+49+y^2+6y+9=49\\\\\Rightarrow x^2+y^2-14x+6y+9=0.$

Thus, the required equation of the circle is $x^2+y^2-14x+6y+9=0.$

Answer 2

(x – h)² + (y – k)² = r² 

Fill in the numbers 

(x – 7)² + (y +3)² = 49