MHB Calculating Radius of Circular Section from Sphere and Plane Intersection

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To find the radius of the circular section formed by the intersection of the sphere defined by x^2 + y^2 + z^2 = 49 and the plane 2x + 3y - z - 5√14 = 0, one approach is to calculate the distance from the sphere's center to the plane. This distance can be determined using the formula d = |Ax0 + By0 + Cz0 + D| / √(A^2 + B^2 + C^2). Once the distance d is found, the radius r of the circular section can be derived using the right triangle formed by the sphere's radius and the distance to the plane. This method simplifies the problem without needing to substitute z back into the sphere's equation. The final radius of the circular section can then be calculated effectively.
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Find the radius of the circular section of the sphere of the sphere x^2 + y^2 + z^2 = 49 by the plane 2x+3y-z-5 \sqrt{14}= 0
 
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Suppose you solve the equation of the plane for $z$, and then substitute for $z$ into the equation of the sphere...what do you get?
 
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I think it is easier to find the distance $d$ from the center of the sphere to the plane (recall the the distance from $(x_0,y_0,z_0)$ to $Ax+By+Cz+D=0$ is $\dfrac{Ax_0+By_y+Cz_0+D}{\sqrt{A^2+B^2+C^2}}$) and then find the radius $r$ of the required circular section from the right triangle.
 

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