Question

The equation of a circle is 2x²+2y²-6x-8=0. The line 3x+4y=17 and another tangent to the circle at Q((7)/(2),-(3)/(2)) intersect at a point P.

Answer 1

To find the point of intersection, solve the system of equations formed by the line and the equation of the circle simultaneously.

Step-by-step explanation:

To find the point of intersection, we can solve the system of equations formed by the line and the equation of the circle simultaneously. First, let's rearrange the equation of the circle by completing the square:

2x² - 6x + 2y² - 8 = 0

x² - 3x + y² - 4 = 0

(x² - 3x + 9/4) + (y² - 4) = 9/4 + 4

(x - 3/2)² + (y - 2)² = 41/4

Now, substitute the equation of the line into the equation of the circle:

(3x + 4y - 17)² - (x - 7/2)² - (y + 3/2)² = 0

Simplifying and combining like terms:

10x + 12y - 115 = 0

Now, solve the system of equations to find the point of intersection, P.