Answer:
\((x-2)^2 +(y-4)^2=25\)
Step-by-step explanation:
Since, the diameter of the circle having the end points (-2, 1) and (6, 7),
Thus, by the distance formula,
\(\text{Diameter}=\sqrt{(6+2)^2+(7-1)^2}\)
\(=\sqrt{8^2+6^2}\)
\(=\sqrt{64+36}\)
\(=\sqrt{100}\)
\(=10\text{ unit}\)
So, the radius of the circle = \(\frac{10}{2}\) = 5 unit,
Let the equation of the circle is,
\((x-h)^2+(y-k)^2=25\)
Where, (h, k) is the center of the circle,
Since, points (-2, 1) and (6, 7) are on the circle,
They must satisfy the equation of the circle,
\((-2-h)^2+(1-k)^2 = 25\)
\((6-h)^2+(7-k)^2=25\)
By solving these equations,
We get,
h = 2, k = 4
Hence, the equation of the circle would be,
\((x-2)^2 +(y-4)^2=25\)
