Mathematics · Circles
JEE Main 2026 — 8 April, Evening Shift — Question 41
Consider the circle C : . Let a variable chord AB of the circle C subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord AB is the circle , then is equal to
Answer: 18
Numerical answer — enter this value.
Step-by-step solution
of chord AB
\mathrm{hx}+\mathrm{ky}=\mathrm{h}^{2}+\mathrm{k}^{2} \end{gathered}$$ Homogenising the eq ${ }^{\mathrm{n}}$ of the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}\left(\frac{\mathrm{hx}+\mathrm{ky}}{\mathrm{h}^{2}+\mathrm{k}^{2}}\right)-8 \mathrm{y}\left(\frac{\mathrm{hx}+\mathrm{ky}}{\mathrm{h}^{2}+\mathrm{k}^{2}}\right)-11\left(\frac{\mathrm{hx}+\mathrm{ky}}{\mathrm{h}^{2}+\mathrm{k}^{2}}\right)^{2}=0$ coeff. of $x^{2}+$ coeff of $y^{2}=0$ $1-\frac{6 \mathrm{~h}}{\mathrm{~h}^{2}+\mathrm{k}^{2}}-11 \cdot \frac{\mathrm{~h}^{2}}{\left(\mathrm{~h}^{2}+\mathrm{k}^{2}\right)^{2}}+1-\frac{8 \mathrm{k}}{\mathrm{h}^{2}+\mathrm{k}^{2}}-11 \cdot \frac{\mathrm{k}^{2}}{\left(\mathrm{~h}^{2}+\mathrm{k}^{2}\right)^{2}}=0$ $2\left(h^{2}+k^{2}\right)-6 h-8 k-11=0$ $\mathrm{x}^{2}+\mathrm{y}^{2}-3 \mathrm{x}-4 \mathrm{y}-\frac{11}{2}=0$ $\therefore \alpha+\beta+2 \gamma=3+4+11=18$
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2026
- Subject
- Mathematics
- Chapter
- Circles
- Topic
- Introduction to Circles