NCERT · Class 11 · Mathematics · Conic SectionsThe radius of the circle $x^2 + y^2 + 8x - 6y + 9 = 0$ is:
Step-by-Step Solution
Comparing the given equation with the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$, we get $2g = 8 \Rightarrow g = 4$, $2f = -6 \Rightarrow f = -3$, and $c = 9$. The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4^2 + (-3)^2 - 9} = \sqrt{16 + 9 - 9} = \sqrt{16} = 4$.
Detailed Options Breakdown
Option : $3$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Conic Sections.
Option 1: $4$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 2: $5$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Conic Sections.
Option 3: $9$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Conic Sections.
💡 Study Guide: This question tests core syllabus concepts from Conic Sections. For formulas, key summaries, and mock exam reference guides, read the full Conic Sections Revision Notes.