Find the equation of the tangent to the circle (x-3)^2 + (y-4)^2 = 13 that passes through the point (1,7)

Start with a sketch. We can see that the radius from the point (1,7) to the centre of the circle (3,4) is perpendicular to the tangent. The gradient of the radius is (4-7)/(3-1) = -3/2. We know that two perpendicular gradients multiply to make -1, so the gradient, m, of the tangent is 2/3.
The equation of the tangent is now y=2/3x + c . To find c, all we have to do is plug in a coordiante for x and y - we know (1,7) lies on the tangent so we will use this. Therefore c=19/3. The equation of our tangent is therefore y=2/3 x + 19/3 !

Answered by Maths tutor

3206 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I find and determine the nature of stationary points of a function?


What are the limits of an inverse tan graph.


Given that (cos(x)^2 + 4 sin(x)^2)/(1-sin(x)^2) = 7, show that tan(x)^2 = 3/2


Solve the following integral: ∫ arcsin(x)/sqrt(1-x^2) dx


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