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Coordinate Geometry
EXERCISE 7.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Step-by-Step Solution

Key Concept: For a diameter AB of a circle, the centre of the circle is the midpoint of AB. Use the midpoint formula \(M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\) and equate it to the given centre.
1. Let the required point be \(A(x_1, y_1)\) and the given point be \(B(1, 4)\).\
2. Since AB is a diameter, the centre \(C(2, -3)\) is the midpoint of AB.\
3. Apply the midpoint formula:
\[\left(\frac{x_1+1}{2},\frac{y_1+4}{2}\right) = (2, -3)\]\
4. Equate the corresponding coordinates:
\[\frac{x_1+1}{2}=2 \quad\text{and}\quad \frac{y_1+4}{2}=-3\]\
5. Solve each equation:
- From \(\frac{x_1+1}{2}=2\): \(x_1+1 = 4 \Rightarrow x_1 = 3\).\
- From \(\frac{y_1+4}{2}=-3\): \(y_1+4 = -6 \Rightarrow y_1 = -10\).\
6. Hence the coordinates of point \(A\) are \((3, -10)\).

Correct Answer: (3, -10)
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