A circle passing through the point P(α ,β ) in the first quadrant touches the two coordinate axes at the points A and B. The point P is above the…

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

$(x-r{)}^{2}+(y-r{)}^{2}={r}^{2}$(x−r)2+(y−r)2=r2 which passes through $(\alpha ,\beta )$(α,β)

So, $(\alpha -r{)}^{2}+(\beta -r{)}^{2}={r}^{2}.......(1)$(α−r)2+(β−r)2=r2.......(1)

Also equation of line $AB:x+y=r$AB:x+y=r

Let $Q(h,k)$Q(h,k),

Now using the formula of foot of perpendicular,

$$\frac{h-\alpha }{1}=\frac{k-\beta }{1}=\frac{-(\alpha +\beta -r)}{2}$$1h−α​=1k−β​=2−(α+β−r)​

$$\Rightarrow (h,k)=(\alpha -\frac{\alpha }{2}-\frac{\beta }{2}+\frac{r}{2},\beta -\frac{\alpha }{2}-\frac{\beta }{2}+\frac{r}{2})$$⇒(h,k)=(α−2α​−2β​+2r​,β−2α​−2β​+2r​)

$$\Rightarrow (h,k)=(\frac{\alpha -\beta +r}{2},\frac{-\alpha +\beta +r}{2})$$⇒(h,k)=(2α−β+r​,2−α+β+r​)

Now using the distance formula we get,

$${(PQ)}^{2}={(\frac{-\alpha -\beta +r}{2})}^{2}+{(\frac{-\alpha -\beta +r}{2})}^{2}=121$$(PQ)2=(2−α−β+r​)2+(2−α−β+r​)2=121

$$\Rightarrow (r-(\alpha +\beta ){)}^{2}=242$$⇒(r−(α+β))2=242

$$\Rightarrow {r}^{2}+(\alpha +\beta {)}^{2}-2r(\alpha +\beta )=242$$⇒r2+(α+β)2−2r(α+β)=242

$$\Rightarrow {\alpha }^{2}+{\beta }^{2}+2\alpha \beta +{r}^{2}-2r\alpha -2r\beta =242$$⇒α2+β2+2αβ+r2−2rα−2rβ=242

$\Rightarrow 2\alpha \beta =242$⇒2αβ=242 ${\text{from equation}(1),(\alpha -r{)}^{2}+(\beta -r{)}^{2}={r}^{2}}$from equation(1),(α−r)2+(β−r)2=r2

$$\Rightarrow \alpha \beta =121$$⇒αβ=121