Question:
If the centre of the circle inscribed in a square formed by the lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ is $(a,b)$, then $a + b$ is
If the centre of the circle inscribed in a square formed by the lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ is $(a,b)$, then $a + b$ is
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When quadratic equations define boundaries, factor them to get straight lines and then find the midpoint to locate the center.
Updated On: Apr 30, 2026
- $11$
- $9$
- $7$
- $5$
- $4$
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The Correct Option is B
Solution and Explanation
Concept:
A square formed by vertical and horizontal lines will have its center at the midpoint of the boundary lines. The inscribed circle shares the same center.
Step 1: Solve given equations to find boundary lines.
For $x$: \[ x^2 - 8x + 12 = 0 \] \[ (x-2)(x-6) = 0 \] \[ x = 2, x = 6 \] For $y$: \[ y^2 - 14y + 45 = 0 \] \[ (y-5)(y-9) = 0 \] \[ y = 5, y = 9 \]
Step 2: Identify square boundaries.
The square is bounded by: \[ x = 2, x = 6, y = 5, y = 9 \]
Step 3: Find center of the square.
\[ a = \frac{2 + 6}{2} = 4 \] \[ b = \frac{5 + 9}{2} = 7 \]
Step 4: Compute required value.
\[ a + b = 4 + 7 = 11 \]
Final Answer: \[ \boxed{11} \]
A square formed by vertical and horizontal lines will have its center at the midpoint of the boundary lines. The inscribed circle shares the same center.
Step 1: Solve given equations to find boundary lines.
For $x$: \[ x^2 - 8x + 12 = 0 \] \[ (x-2)(x-6) = 0 \] \[ x = 2, x = 6 \] For $y$: \[ y^2 - 14y + 45 = 0 \] \[ (y-5)(y-9) = 0 \] \[ y = 5, y = 9 \]
Step 2: Identify square boundaries.
The square is bounded by: \[ x = 2, x = 6, y = 5, y = 9 \]
Step 3: Find center of the square.
\[ a = \frac{2 + 6}{2} = 4 \] \[ b = \frac{5 + 9}{2} = 7 \]
Step 4: Compute required value.
\[ a + b = 4 + 7 = 11 \]
Final Answer: \[ \boxed{11} \]
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