A circle has equation: (x - 2)^2 + (y - 2)^2 = 16. It intersects the y-axis (y > 0) at point P and the x-axis (x < 0) at point Q. Find the equation of the line connecting P and Q and of the line perpendicular to PQ passing through the circle's centre.

(x - 2)2 + (y - 2)2 = 16 x = 0 --> 4 + (y - 2)2 = 16 --> y - 2 = ±2sqrt{2} --> y = 2 ± 2sqrt{2} --> P(0,2 + 2sqrt{2})y = 0 --> (x - 2)2 + 4 = 16 --> x - 2 = ±2sqrt{2} --> x = 2 ± 2sqrt{2} --> Q(2 - 2sqrt{2},0) Line PQ: Gradient = (0 - (2 + 2sqrt{2}))/((2 - 2sqrt{2}) - 0) = 3 + 2sqrt{2} Equation of line PQ: y = (3 + 2sqrt{2})x + (2 + 2sqrt{2}) Line perpendicular to PQ passing through the circle's centre: Gradient = -1/(3 + 2sqrt{2}) = 2sqrt{2} - 3 Equation of line: y - 2 = (2sqrt{2} - 3)(x - 2) --> y = (2sqrt{2} - 3)x + (8 - 4sqrt{2})

Answered by Maths tutor

3052 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A circle with centre C(2, 3) passes through the point A(-4,-5). (a) Find the equation of the circle in the form (x-a)^2 + (y-b)^2=k


How do I deal with parametric equations? x = 4 cos ( t + pi/6), y = 2 sin t, Show that x + y = 2sqrt(3) cos t.


Express the fraction (p+q)/(p-q) in the form m+n√2, where p=3-2√2 and q=2-√2.


Solve the equation 3 sin^2 theta = 4 cos theta − 1 for 0 ≤ theta ≤ 360


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning