Question

The endpoints of a diameter of a circle are (6,2) and (−2,5) . What is the standard form of the equation of this circle?

Answer 1

Answer:

The standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$

Step-by-step explanation:

Given that the end points of a diameter of a circle are (6,2) and (-2,5);

Now to find the standard form of the equation of this circle:

The center is (h,k) of the circle is the midpoint of the given diameter

midpoint formula is $M=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

Let $(x_{1},y_{1})$ and $(x_{2},y_{2})$ be the given points (6,2) and (-2,5) respectively.

$M=(\frac{6-2}{2},\frac{2+5}{2})$

$M=(\frac{4}{2},\frac{7}{2})$

$M=(2,\frac{7}{2})$

Therefore the center (h,k) is $(2,\frac{7}{2})$

now to find the radius:

The diameter is the distance between the given points (6,2) and (-2,5)

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$=\sqrt{(-2-6)^2+(5-2)^2}$

$=\sqrt{(-8)^2+(3)^2}$

$=\sqrt{64+9}$

$=\sqrt{73}$

Therefore the radius is $\sqrt{73}$

i.e., $r=\sqrt{73}$

Therefore the standard form of the circle is

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

Now substituting the center  and radiuswe get

$(x-2)^2+(y-\frac{7}{2})^2=(\sqrt{73})^2$

$(x-2)^2+(y-\frac{7}{2})^2=73$

Therefore the standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$