Math Is Fun Forum

Rinni

Member

Registered: 2014-10-31

Posts: 16

circle

1. if the line 2x-y+1=0  touches the circle at the point (2,5) and the centre of the circle lies on the line x+y-9=0. find the equation of the circle.

2.the variable coefficients a,b in the equation of the straight line x/a+y/b=1 are connected by the relation 1/a^2+1/b^2=1/c^2 where c is a fixed constant. show that the locus of the foot of the perpendicular from the origin upon the line is a circle. find the equation of the circle

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Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: circle

hi Rinni,

Welcome to the forum.

So there's a tangent to the circle at (2,5).  You know its gradient so you can make the equation of the radius line at that point (gradient = -1/m of tangent)

Then find where it intersects the other line and you've got the centre.

Got to go out for a while.  I'll think about the second one and come back here later.  If you read the above in the meantime, I suggest you post your working so I can check it for you. 

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 24,996

Website

Re: circle

I'll assume that by touch, you mean that it is a tangent.

Let (a,b) be the center of the circle and r be its radius

a + b - 9 = 0 ... (i)

r^2 = (2-a)^2 + (5-b)^2 = (2 a - b + 1)^2/5 ... (ii)

Solving (i) and (ii), a = 6, b = 3, r^2 = 20

So, the circle is (x - 6)^2 + (y - 3)^2 == 20€

Last edited by Agnishom (2014-10-31 23:24:04)

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'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: circle

hi Rinni,

Here's how to attempt (2).

You know

So you can get the gradient and the equation of the perpendicular in the form y = mx where m depends on a and b.

They intersect on the locus so solve as simultaneous equations for x and y.

Also you know

Rearrange this to get it in the form c^2 = ???? in terms of a and b.

As you have the hint that it's going to come out as a circle, right down expressions for x^2 and for y^2 and add them together.

Use the above to replace all the a and b expressions with a single one for c and it's done.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Rinni

Member

Registered: 2014-10-31

Posts: 16

Re: circle

sorry Bob, I checked it now only........... doing the 1st one and thanks for the second sum.

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"keep smiling , it costs nothing "