AP DSC School Assistant Math Model Test 2

Concept:The point where a plane touches a sphere (point of tangency) lies on both the sphere and the plane. The equation of the tangent plane to a sphere at a given point can be derived from the sphere’s equation; comparing it with the given plane yields the coordinates of the point.Explanation:Step 1: Write the sphere equation in standard form.The given sphere is$x^2+y^2+z^2-2x-4y+2z-3=0$. Comparing with$x^2+y^2+z^2+2ux+2vy+2wz+d=0$, we get$u=-1$,$v=-2$,$w=1$,$d=-3$.Step 2: Equation of the tangent plane at a point$P(a,b,c)$on the sphere is:$ax+by+cz + u(x+a) + v(y+b) + w(z+c) + d = 0$.Substituting values:$ax+by+cz -1(x+a) -2(y+b) +1(z+c) -3 = 0$Simplify:$(a-1)x + (b-2)y + (c+1)z + (c - a - 2b - 3) = 0$.Step 3: The given tangent plane is$2x - 2y + z + 12 = 0$. Since both represent the same plane, their coefficients are proportional:$\frac{a-1}{2} = \frac{b-2}{-2} = \frac{c+1}{1} = \frac{c-a-2b-3}{12}$.Step 4: Let the common ratio be$k$. Then:$a-1 = 2k \Rightarrow a = 2k+1$$b-2 = -2k \Rightarrow b = -2k+2$$c+1 = k \Rightarrow c = k-1$Step 5: Substitute into the constant term ratio:$\frac{c-a-2b-3}{12} = k$$\frac{(k-1) - (2k+1) - 2(-2k+2) - 3}{12} = k$Compute numerator:$k-1-2k-1+4k-4-3 = 3k-9$So$\frac{3k-9}{12}=k \Rightarrow 3k-9=12k \Rightarrow -9=9k \Rightarrow k=-1$.Step 6: Find coordinates:$a = 2(-1)+1 = -1$$b = -2(-1)+2 = 2+2 = 4$$c = -1-1 = -2$Thus the point is$(-1,4,-2)$.Step 7: Verify that this point satisfies the plane equation:$2(-1) - 2(4) + (-2) + 12 = -2-8-2+12 = 0$. It also satisfies the sphere equation (not shown but holds true). Therefore it is the correct point of tangency. Answer:The point of contact is$(-1, 4, -2)$.