Question

What is the equation of the circle with center (3.2,-2.1) and radius 4.3?

Answer 1

(x-3.2)^2 +(y+2.1)=4.3^2

Answer 2

Answer:  The required equation of the circle is $x^2+y^2+6.4x+4.2y=3.84.$

Step-by-step explanation:  We are given to find the equation of the circle with center (3.2,-2.1) and radius 4.3.

We know that

the STANDARD equation of a circle with center (h, k) and radius r units is given by

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

For the given circle,

center, (h, k) = (3.2, -2.1)  and radius, r = 4.3 units.

So, from equation (i), we get

$(x-3.2)^2+(y-(-2.1))^2=(4.3)^2\\\\\Rightarrow (x-3.2)^2+(y+2.1)^2=18.49\\\\\Rightarrow x^2-6.4x+10.24+y^2+4.2y+4.41=18.49\\\\\Rightarrow x^2+y^2-6.4x+4.2y+14.65=18.49\\\\\Rightarrow x^2+y^2+6.4x+4.2y=18.49-14.65\\\\\Rightarrow x^2+y^2+6.4x+4.2y=3.84.$

Thus, the required equation of the circle is $x^2+y^2+6.4x+4.2y=3.84.$