Question

P 18 The circle C has equation (x-3)² + (y + 3)² = 52. The baselines, and are tangents to the circle and have gradient a Find the points of intersection, P and Q, of the tangents and the circle. b Find the equations of lines and 12, giving your answers in the form ax + by + c = 0.​

Answer 1

To solve the student's question, we would identify the circle's center and radius from its equation, use the gradient 'a' to find the points of tangency, and then derive the tangent lines' equations. The answer requires additional information about the gradient.

Step-by-step explanation:

The question involves finding points of intersection between tangents and a circle, and then deriving the equations of those tangent lines. First, we need to recognize that the equation of the circle (x-3)² + (y+3)² = 52 determines a circle centered at (3, -3) with a radius of √52.

To find the points of intersection (P and Q), we would typically set the equations of the tangents (which involve the slope 'a') equal to the equation of the circle and solve for the x and y coordinates. However, the gradient 'a' has not been provided, which makes finding the exact points not possible in this context.

Nevertheless, once we have the points of intersection and the slope 'a', we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point of intersection and 'm' is the slope, to write the equation of the tangent lines. We can then rearrange this into the ax + by + c = 0 format requested.

Answer 2

a) To find the points of intersection, P and Q, of the tangents and the circle, we can use the fact that a tangent line is perpendicular to the radius at the point of tangency. Therefore, the slope of the tangent line is the negative reciprocal of the slope of the radius.

We know that the equation of the circle is (x-3)² + (y + 3)² = 52. To find the slope of the radius, we can differentiate the equation of the circle with respect to x and y.

dx/dt = 2(x-3) and dy/dt = 2(y+3)

We know that the slope of the tangents is a. So, the slope of the radius is -1/a

Now we can use the point slope form of a line to find the equation of the tangent line.

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is a point on the line.

We can substitute the point of tangency, which is on the circle, into the point slope form.

y + 3 = -1/a(x - 3)

We can now substitute the point (3, -3) and the slope -1/a into the point slope form to find the equation of the line

y + 3 = -1/a(x - 3)

y = -1/a

Step-by-step explanation:

Not sure if this helps, hope it does!