Step-by-step explanation:
Let's say the coordinates of Q are (h, k).
The slope of PQ is perpendicular to the tangent line.
(k − 2) / (h − 2) = -⅓
3(k − 2) = -(h − 2)
3(k − 2) = 2 − h
Solving for h:
h = 2 − 3(k − 2)
h = 2 − 3k + 6
h = 8 − 3k
(2, 2) is a point on the circle, so:
(2 − h)² + (2 − k)² = r²
Substituting:
9 (k − 2)² + (2 − k)² = r²
10 (k − 2)² = r²
The equation of the circle is:
(x − 8 + 3k)² + (y − k)² = 10 (k − 2)²
Without more information, there are an infinite number of possible solutions. For example, if k = 1, the equation of the circle is:
(x − 5)² + (y − 1)² = 10
