Question

find the standard form of the equation of the circle with endpoints of a diameter at the points (1,8) and (-9,4). type the standard form of the equation of this circle. (type an equation.)

Explanation:

Step1: Find the center of the circle

The center of the circle is the mid - point of the diameter. The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here, $x_1=1,y_1 = 8,x_2=-9,y_2 = 4$. So the center $(h,k)=(\frac{1+( - 9)}{2},\frac{8 + 4}{2})=(\frac{1-9}{2},\frac{12}{2})=(-4,6)$.

Step2: Find the radius of the circle

The radius $r$ is the distance between the center $(h,k)=(-4,6)$ and one of the endpoints of the diameter, say $(x_1,y_1)=(1,8)$. The distance formula is $r=\sqrt{(x_1 - h)^2+(y_1 - k)^2}$. Substitute the values: $r=\sqrt{(1-( - 4))^2+(8 - 6)^2}=\sqrt{(1 + 4)^2+(8 - 6)^2}=\sqrt{5^2+2^2}=\sqrt{25 + 4}=\sqrt{29}$.

Step3: Write the standard form of the circle equation

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$. Substitute $h=-4,k = 6,r=\sqrt{29}$ into the equation: $(x+4)^2+(y - 6)^2=29$.

Answer:

$(x + 4)^2+(y - 6)^2=29$