To solve the problem, we need to find the centre of a circle when the endpoints of its diameter are given as (1, 2) and (7, –4).
1. Understanding the Concept:
The centre of a circle is the midpoint of the diameter. If the endpoints of the diameter are \( (x_1, y_1) \) and \( (x_2, y_2) \), then the midpoint is given by:
\[
\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
2. Substitute the values:
Given points: \( (1, 2) \) and \( (7, -4) \)
\[
x = \frac{1 + 7}{2} = \frac{8}{2} = 4
\]
\[
y = \frac{2 + (-4)}{2} = \frac{-2}{2} = -1
\]
3. Calculate the Coordinates of the Centre:
The centre of the circle is at \( (4, -1) \).
Final Answer:
The correct answer is option (B): (4, –1).
In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity:
Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)
In the following figure, circle with centre D touches the sides of \(\angle\)ACB at A and B. If \(\angle\)ACB = 52\(^\circ\), find measure of \(\angle\)ADB.