JEE Main 2026 Apr 6 Shift 2, Mathematics Q23: Circle: Chord, Tangent And Normal

Find $Q,R$. Substitute $x=y+4$ into the circle: $y^2+(y+3{)}^2=9$ $\Rightarrow$ $2y^2+6y=0$ $\Rightarrow$ $y=0$ or $-3$. Points: $Q(4,0)$ and $R(1,-3)$.

Conditions on $P$. $P$ lies on circle: $(\alpha -4{)}^2+(\beta +3{)}^2=9$ ...(i). Also $PQ=PR$:

$(\alpha -4{)}^2+{\beta }^2=(\alpha -1{)}^2+(\beta +3{)}^2$ $\Rightarrow$ $-8\alpha +16=-2\alpha +1+6\beta +9$ $\Rightarrow$ $\alpha +\beta =1$ ...(ii)

Solve. From (ii), $\beta =1-\alpha$. Substituting into (i):

$(\alpha -4{)}^2+(4-\alpha {)}^2=9$ $\Rightarrow$ $2(\alpha -4{)}^2=9$ $\Rightarrow$ $\alpha -4=\pm \frac{3}{\sqrt{2}}$.

So $\alpha =4\pm \frac{3}{\sqrt{2}}$ and $\beta =-3\mp \frac{3}{\sqrt{2}}$.

$6\alpha +8\beta =6(4)+8(-3)+(6\mp 8)\cdot \frac{3}{\sqrt{2}}=0+(\mp 2)\cdot \frac{3}{\sqrt{2}}=\mp \frac{6}{\sqrt{2}}=\mp 3\sqrt{2}$.

$(6\alpha +8\beta {)}^2=18$.