Homework Help

Given that a circle passes through the points P(3,5) and Q(-1,3) has a radius square...

user profile pic

sherryseah | Student, Grade 9 | Salutatorian

Posted June 30, 2013 at 5:38 AM via web

dislike 1 like

Given that a circle passes through the points P(3,5) and Q(-1,3) has a radius square root of 10, find the eqn of the circle

1 Answer | Add Yours

user profile pic

aruv | High School Teacher | Valedictorian

Posted June 30, 2013 at 6:09 AM (Answer #1)

dislike 1 like

We have given two points P(3,5) and Q(-1,3). Also radius of the circle `sqrt(10)`  unit.  We wish to dertermine equation of the circle.

Let equation of the circle whose centre at (h,k) and radius r be

`(x-h)^2+(y-k)^2=r^2`      (i)

(i) posses through P and Q ,therefore

`(3-h)^2+(5-k)^2=10`

`` `9+h^2-6h+25+k^2-10k=10`

`h^2+k^2-6h-10k=-24`    (ii)

`(-1-h)^2+(3-k)^2=10`

`1+h^2+2h+9+k^2-6k=10`

`h^2+k^2+2h-6k=0`       (iii)

Solving (ii) and (iii),we have

2h+k=6

k=6-2h     (iv)

put k ,from(iv) in (iii)

`h^2+(6-2h)^2+2h-6(6-2h)=0`

`h^2+36+4h^2-24h+2h-36+12h=0`

`5h^2-10h=0`

`h=0 or h=2`

h=0 then k=6

h=2 then  k=2

Thus there are two circles

`(x-0)^2+(y-6)^2=10`

`x^2+y^2-12y+36=10`

`x^2+y^2-12y+26=0`

and

`(x-2)^2+(y-2)^2=10`

`x^2+4-4x+y^2+4-4y=10`

`x^2+y^2-4x-4y-2=0`

Hence we have two circle.

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes