Let the centre of a circle C be α ,β and its radius r<8. Let 3x+4y=24 and 3x–4y=32 be two tangents and 4x+3y=1 be a normal to C. Then (α -β +r) is…

Given,

The centre of a circle $C$C be $\alpha ,\beta$α,β and its radius $r<8$r<8,

And $3x+4y=24$3x+4y=24 and $3x–4y=32$3x–4y=32 be two tangents

So, the distance from centre will be radius,

$$|\frac{3\alpha +4\beta -24}{5}|=|\frac{3\alpha -4\beta -32}{5}|$$∣53α+4β−24​∣=∣53α−4β−32​∣

Taking $+$+ sign we get,

$$\Rightarrow 3\alpha +4\beta -24=3\alpha -4\beta -32$$⇒3α+4β−24=3α−4β−32

$$\Rightarrow 8\beta =-8\Rightarrow \beta =-1$$⇒8β=−8⇒β=−1

And taking $-$− sign we get,

$$3\alpha +4\beta -24=-3\alpha +4\beta +32$$3α+4β−24=−3α+4β+32

$$\Rightarrow 6\alpha =56\Rightarrow \alpha =\frac{56}{6}$$⇒6α=56⇒α=656​

And $4x+3y=1$4x+3y=1 be a normal to $C$C

So, $(\alpha ,\beta )$(α,β) lies on $4x+3y=1$4x+3y=1

$$\Rightarrow 4\alpha +3\beta =1$$⇒4α+3β=1

So, $\alpha =1$α=1 when $\beta =-1$β=−1 and $\beta =\frac{1}{3}(1-4\times \frac{28}{3})=\frac{-109}{9},\text{if}\alpha =\frac{28}{3}$β=31​(1−4×328​)=9−109​,ifα=328​

Hence, radius will be,

$r=|\frac{3-4-24}{5}|=5<8$r=∣53−4−24​∣=5<8 if $(\alpha ,\beta )=(1,-1)$(α,β)=(1,−1)

And if $(\alpha ,\beta )=(\frac{28}{3},\frac{-109}{9})$(α,β)=(328​,9−109​) then $r>8$r>8, so neglected.

$$\therefore r=5,\alpha =1,\beta =-1$$∴r=5,α=1,β=−1

$$\therefore \alpha -\beta +r=7$$∴α−β+r=7