Find the equation of the tangent to the circle x² + y² = 9, which passes through the point on the circle where x = 2 and y is positive.
Find the equation of the tangent to the circle x² + y² = 9, which passes through the point on the circle where x = 2 and y is positive.
correct answer is: option b : √5y = –2x + 9
Explanation:\(x^2 + y^2 = 9\)
Given the circle equation \(x^2 + y^2 = 9\), we need to find the equation of the tangent at the point where \(x = 2\) and \(y\) is positive.
Step 1: Find the point of tangency
Substitute \(x = 2\) into the circle equation to find \(y\):
\(2^2 + y^2 = 9 \implies 4 + y^2 = 9 \implies y^2 = 5 \implies y = \sqrt{5}\) (since \(y\) is positive)
Thus, the point of tangency is \((2, \sqrt{5})\)
Step 2: Use the tangent line formula
The equation of the tangent to the circle at point \((x_1, y_1)\) is given by:
\(xx_1 + yy_1 = r^2\)
Substituting \((x_1, y_1) = (2, \sqrt{5})\) and \(r = 3\):
\(2x + \sqrt{5}y = 9\)
Step 3: Match with the given options
Rearrange the equation to match the options:
\(\sqrt{5}y = -2x + 9\)
This corresponds to option b: \(\sqrt{5}y = -2x + 9\)
Final Answer: \(\boxed{\sqrt{5}y = -2x + 9}\)
