The circle $x^2+y^2-8 x-12 y+\alpha=0$ lies in the first quadrant without touching the coordinate axes. If…

The circle $x^2+y^2-8 x-12 y+\alpha=0$ lies in the first quadrant without touching the coordinate axes. If $(6,6)$ is an interior point to the circle, then
  1. $4 \lt \alpha \lt 6$
  2. $6 \lt \alpha \lt 16$
  3. $16 \lt \alpha \lt 48$
  4. $36 \lt \alpha \lt 48$

Solution

$(6,6)$ is an interior point of $x^2+y^2-8 x-12 y+\alpha=0$ $\Rightarrow 36+36-48-72+\alpha \lt 0 \Rightarrow \alpha \lt 48$
Also, radius $ \lt x$ and $y$ coordinate of centre $\Rightarrow \sqrt{16+36-\alpha} \lt 4$ and 6 $\Rightarrow 52-\alpha \lt 16$ and $36 \Rightarrow \alpha \geqslant 36 \therefore 36 \lt \alpha \lt 48$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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