The equation of a circle is x^2+y^2-6x-4y+4=0. i) Find the radius and centre of the circle. ii) Find the coordinates of the points of intersection with the line y=x+2

i) x2 + y2- 6x - 4y + 4 = 0 x2- 6x + y2 - 4y + 4 = 0 group the terms together x2- 6x = (x - 3)2 - 9 factorise the x terms y2 - 4y = (y - 2)2 - 4 factorise the y terms x2- 6x + y2 - 4y + 4 = (x - 3)2 - 9 + (y - 2)2 - 4 + 4 = 0 put them together(x - 3)2 + (y - 2)2 = 9 put constant terms on right side of equationCentre: (3,2) radius: 3
ii) x2 + (x + 2)2- 6x - 4(x + 2) + 4 = 0 substitute y = x + 2 into the original equation x2 + x2 + 4x + 4 - 6x - 4x - 8 + 4 = 0 expand the brackets2x2 - 6x = 0 simplify expression (cancel terms) x2 - 3x = 0 simplify expression (divide by 2)x(x - 3) = 0 factorise equationx = 0 or 3 solve for xwhen x = 0, y = x + 2 = 2 substitute x values into line equation to get y valueswhen x = 3, y = x + 2 = 5points of intersection are (0, 2) and (3, 5)

Answered by Ralf L. Maths tutor

6081 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do we differentiate y=a^x when 'a' is an non zero real number


a) show that (cosx)^2=8(sinx)^2-6sinx can be written as (3sinx-1)^2=2 b)Solve (cosx)^2=8(sinx)^2-6sinx


A curve C with an equation y = sin(x)/e^(2x) , 0<x<pi has a stationary point at P. Find the coordinates ofP?


Show that 2sin(2x)-3cos(2x)-3sin(x)+3=sin(x)(4cos(x)+6sin(x)-3)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy