A circle passing through the point P(α, β) in the first quadrant...

Let the equation of circle touching both the coordinate axis be,

$( x - r{)}^{2} + ( y - r{)}^{2} = {r}^{2}$ which passes through \left(\alpha, \beta\right)$</p><p>So,$( \alpha - r{)}^{2} + ( \beta - r{)}^{2} = {r}^{2} . . . . . . . \left(1\right)$</p><p>Also equation of line$AB: x + y = r$</p><p> </p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/744d7ab7-d7e1-438e-b442-2d90dd9a532e-image.png" style="width: 200px; height: 166px;" /></p><p>Let$Q\left(h, k\right)$,</p><p>Now using the formula of foot of perpendicular,</p><p>$\dfrac{h - \alpha}{1} = \dfrac{k - \beta}{1} = \dfrac{- ( \alpha + \beta - r )}{2}$</p><p>$⇒ ( h, k ) = \left(\alpha - \dfrac{\alpha}{2} - \dfrac{\beta}{2} + \dfrac{r}{2}, \beta - \dfrac{\alpha}{2} - \dfrac{\beta}{2} + \dfrac{r}{2}\right)$</p><p>$⇒ \left(h, k\right) = \left(\dfrac{\alpha - \beta + r}{2}, \dfrac{- \alpha + \beta + r}{2}\right)$</p><p>Now using the distance formula we get,</p><p>${\left(PQ\right)}^{2} = {\left(\dfrac{- \alpha - \beta + r}{2}\right)}^{2} + {\left(\dfrac{- \alpha - \beta + r}{2}\right)}^{2} = 121$</p><p>$⇒ ( r - ( \alpha + \beta ) {)}^{2} = 242$</p><p>$⇒ {r}^{2} + ( \alpha + \beta{)}^{2} - 2r ( \alpha + \beta ) = 242$</p><p>$⇒ {\alpha}^{2} + {\beta}^{2} + 2\alpha\beta + {r}^{2} - 2r\alpha - 2r\beta = 242$</p><p>$⇒ 2\alpha\beta = 242$\left\{\text{from equation}\left(1\right), ( \alpha - r{)}^{2} + ( \beta - r{)}^{2} = {r}^{2}\right\}$

$⇒ \alpha\beta = 121$