Co-ordinate Geometry-Cirlcles

• Jul 22nd 2013, 10:00 PM
smatik
Co-ordinate Geometry-Cirlcles
Through a fixed point (h,k) secants are drawn to the circle x2 + y2 =a2. The locus of the mid points of the secants intercepted by the given circle is ....??

My Answer: x2 + y2 = 2(hx + ky)
Book's Answer : x2 + y2 = (hx + ky)

Here's what i did:
Attachment 28884
Maybe the book is correct but i can't figure out what wrong I did.(Sweating)
• Jul 23rd 2013, 06:06 AM
BobP
Re: Co-ordinate Geometry-Cirlcles
Can you begin by explaining how you know, (without seeing the answer to the question) , that the path of the mid-point will be a circle ?
Can you also explain why you think that the radius of that circle will be sqrt(h^2 + k^2) ?

n.b., I've been through the question and arrived at the book answer.
• Jul 23rd 2013, 10:54 AM
smatik
Re: Co-ordinate Geometry-Cirlcles
well if you sketch the locus it looks like a circle[A circle with center at (h,k) and passing through (0,0){since (0,0) will be the mid point of largest secant i.e the diameter of given circle}].Also this was an objective type question(time limit = 2 min)with four and the four options were representing equations of circle..anyway what u did??I used the distance formula...The distance between center of given circle(0,0) and center of required circle(h,k) will be the radius!!!! Please reply with what you did
• Jul 23rd 2013, 02:47 PM
BobP
Re: Co-ordinate Geometry-Cirlcles
There are several points arising from what you say.

First, just because a sketch looks something like part of a circle, it doesn't follow, and you must not assume that, the curve actually is part of a circle. Second, even if the locus is part of a circle you cannot assume that the centre is at (h,k), it isn't. The locus certainly passes through the origin, but since you don't have co-ordinates for the centre you can't calculate the radius.
Simply, it's wrong to make assumptions like this.

What I did for my solution was to take the equation of the variable straight line passing through (h, k) to be

$\frac{y-k}{x-h}=m,$

where $m$ is a parameter, the variable slope of the line.
I calculated the co-ordinates of the points of intersection of the line with the circle and hence the co-ordinates of the centre of the secant. I then eliminated the parameter to obtain the equation of the curve. This, as I said earlier, produced the book answer.

I was somewhat surprised to see that you were allowed just two minutes to answer the question, (it took me about five minutes), but if it's the case that you were given a number of options and just asked to choose, then depending on what the options were, I suppose that two minutes might be long enough.
• Jul 24th 2013, 09:21 PM
smatik
Re: Co-ordinate Geometry-Cirlcles
yeah my assumption that h,k is the centre was wrong.. :P.So your solution is much satisfactory... Mine wass full of assumptions