Mathematics · Circles
JEE Main 2026 — 6 April, Evening Shift — Question 42
Let the line intersect the circle at the points and If is a point on such that then is equal to ______.
Answer: 18
Numerical answer — enter this value.
Step-by-step solution
(x-4)^{2}+(y+3)^{2}=9 \end{gathered}$$ $(\alpha, \beta)$ lies on circle $$\begin{gathered} \therefore(\alpha-4)^{2}+(\beta+3)^{2}=9 \end{gathered}$$ eq. (1) \& (2) $\Rightarrow y^{2}+(y+3)^{2}=9 \Rightarrow 2 y^{2}+6 y=0$ $\Rightarrow \mathrm{y}=0,-3$ $\therefore \mathrm{x}=4,1$ $\mathrm{Q}(4,0), \mathrm{R}(1,-3)$ now $(\mathrm{PQ})^{2}=(\mathrm{PR})^{2}$ $\Rightarrow(\alpha-4)^{2}+\beta^{2}=(\alpha-1)^{2}+(\beta+3)^{2}$ $$\begin{gathered} \Rightarrow \alpha+\beta=1 \end{gathered}$$ eq. (3) \& (4) $\Rightarrow \alpha=4+\frac{3}{\sqrt{2}}, \beta=-3-\frac{3}{\sqrt{2}}$ $\therefore(6 \alpha+8 \beta)^{2}=18$
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2026
- Subject
- Mathematics
- Chapter
- Circles
- Topic
- Considering a Line or a Point wrt a Circle