Question 1140775
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The answer to part (c) from the other tutor is not for the question that is asked....<br>
{{{x^2-6x+y^2-2y=6}}}<br>
Complete the square in x and y to put the equation in standard form<br>
{{{(x-h)^2+(y-k)^2 = r^2}}}<br>
{{{(x^2-6x)+(y^2-2y) = 6}}}
{{{(x^2-6x+9)+(y^2-2y+1) = 6+9+1}}}
{{{(x-3)^2 + (y-1)^2 = 4^2}}}<br>
The center is (3,1); the radius is 4.<br>
We are done with parts (a) and (b) (done correctly by the other tutor).<br>
(c) A tangent PQ is drawn from the point  P(6,-2) to the circle, with Q the point of contact.<br>
Consider the line determined by M and Q.  Let the diameter of the circle contained in that line be AB, with B between M and Q.  Then the rule relating the lengths of a tangent and a secant to a circle tell us<br>
{{{PQ^2 = (PM-MB)(PM+MA)}}}<br>
MA and MB are radii of the circle; length 4.  MP by the Pythagorean Theorem is 3*sqrt(3).  Then<br>
{{{PQ^2 = (3sqrt(3)-4)(3sqrt(3)+4) = 27-16 = 11}}}<br>
ANSWER: PQ = sqrt(11)