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

2 Answers

1 vote
(x-3.2)^2 +(y+2.1)=4.3^2
User Daniel Calliess
by
7.7k points
5 votes

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.

User Sylvanaar
by
8.4k points

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