28. केन्द्र \( (-1,2) \) र व्यास \( 2 \sqrt{5} \) एकाइ मएको एडटा वत्तको अनुसा बिन्दु \( \mathrm{B}(2,3) \) को उत्कम बिन्दु पत्ता लगाडनहासी 1 (Find the igversion point of point \( \mathrm{B}(2,3) \) with respect to circle having centre \( (-1,2) \) and diameter \( 2 \sqrt{5} \) units.) AsM \( B\left(\frac{1}{2}, \frac{5}{2}\right) \)

**Step 1. Identify the Given Information** The circle has centre \( A(-1,2) \) and diameter \( 2\sqrt{5} \). Thus, the radius is \[ r = \frac{2\sqrt{5}}{2} = \sqrt{5}. \] We are given the point \[ B(2,3). \] **Step 2. Compute the Distance from the Centre to the Point \( B \)** Let \[ \overrightarrow{AB} = B - A = (2 - (-1),\, 3 - 2) = (3, 1). \] The distance \( |AB| \) is \[ |AB| = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}. \] **Step 3. Use the Inversion Formula** For the inversion of a point \( B \) with respect to a circle with centre \( A \) and radius \( r \), the inversion point \( B' \) lies on the line through \( A \) and \( B \) and satisfies \[ |AB|\cdot |AB'| = r^2. \] Thus, we have \[ |AB'| = \frac{r^2}{|AB|} = \frac{(\sqrt{5})^2}{\sqrt{10}} = \frac{5}{\sqrt{10}} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2}. \] **Step 4. Determine the Coordinates of \( B' \)** Since \( A \), \( B \), and \( B' \) are collinear, we can express \( B' \) as \[ B' = A + \lambda\, \overrightarrow{AB}, \] with some scalar \( \lambda \). The distance \( |AB'| \) is also given by \[ |AB'| = \lambda\, |AB|. \] Hence, \[ \lambda\, |AB| = \frac{\sqrt{10}}{2} \quad \Longrightarrow \quad \lambda = \frac{\sqrt{10}}{2\sqrt{10}} = \frac{1}{2}. \] Now, substitute \( \lambda = \frac{1}{2} \): \[ B' = \left(-1, 2\right) + \frac{1}{2}(3, 1) = \left(-1+\frac{3}{2},\, 2+\frac{1}{2}\right) = \left(\frac{-2+3}{2},\, \frac{4+1}{2}\right) = \left(\frac{1}{2},\, \frac{5}{2}\right). \] **Step 5. Conclusion** The inversion of the point \( B(2,3) \) with respect to the circle is \[ B'\left(\frac{1}{2}, \frac{5}{2}\right). \]