The circle x^2+y^2-8x-12y+alpha=0 lies in the | vedclass

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(D) The equation of the circle is $x^2+y^2-8x-12y+\alpha=0$. The center is $(4, 6)$ and the radius $r = \sqrt{4^2+6^2-\alpha} = \sqrt{52-\alpha}$.Sinc

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$36 < \alpha < 48$

Vedclass · Answer

(D) The equation of the circle is $x^2+y^2-8x-12y+\alpha=0$. The center is $(4, 6)$ and the radius $r = \sqrt{4^2+6^2-\alpha} = \sqrt{52-\alpha}$.Since the circle lies in the first quadrant without touching the axes,the distance from the center to the axes must be greater than the radius: $r  36$.Since $(6, 6)$ is an interior point,substituting it into the circle equation gives $6^2+6^2-8(6)-12(6)+\alpha $36+36-48-72+\alpha Combining these,we get $36 < \alpha < 48$.

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

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