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secondary 3 | A Maths
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question : find the equation of the circle which passes through the points A(5,0) and B(3,4) , and has its centre lying on the line 2y+x=4
can someone explain what is going on in this question as i dont really understand what is going on here, thx:)
Date Posted: 3 years ago
Views: 192
Wait this one I HAVE TO explain. The concept is tough.
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To find the equation of the circle, we need these two things.
1. Centre
2. Radius
We do not have the radius or the diameter of the circle, so we would need to find the straight-line distance between the centre of the circle and a random point on the circle.
So, in any case, we need to obtain our centre of the circle first, or else we won't even be able to find the radius of the circle. But how do we go about finding the centre of the circle?
-----------------------------------------------------------
It is noteworthy to know that the perpendicular bisector of any chord will pass through the centre of the circle (a perpendicular bisector of another line is a line which divides the other line into two equal parts at right angles).
This means that, once we find the equation of the perpendicular bisector, the centre will lie somewhere along that perpendicular bisector. But this information is not sufficient to obtain the centre of the circle, since the centre can possibly lie anywhere along the line.
A second non-parallel line which also passes the same centre of the circle will now confirm the exact position of the centre of the circle which must be at the intersection of the perpendicular bisector and this second line.
-------------------------------------------------------------
So, our approach is to
1. Find the perpendicular bisector of the particular chord joined by (5, 0) and (3, 4)
--> find midpoint of chord (perpendicular bisector must pass this specific point)
--> find gradient of chord
--> find gradient of perpendicular bisector
--> find the equation of perpendicular bisector
2. Simul our perpendicular bisector with the second equation
3. Solution to the simul is our centre of the circle
4. Use this centre and a point on the circle, namely either (5, 0) or (3, 4), to find our radius
5. Write out the equation of the circle in any form (I prefer standard form because it's more obvious to the eye).
1. Centre
2. Radius
We do not have the radius or the diameter of the circle, so we would need to find the straight-line distance between the centre of the circle and a random point on the circle.
So, in any case, we need to obtain our centre of the circle first, or else we won't even be able to find the radius of the circle. But how do we go about finding the centre of the circle?
-----------------------------------------------------------
It is noteworthy to know that the perpendicular bisector of any chord will pass through the centre of the circle (a perpendicular bisector of another line is a line which divides the other line into two equal parts at right angles).
This means that, once we find the equation of the perpendicular bisector, the centre will lie somewhere along that perpendicular bisector. But this information is not sufficient to obtain the centre of the circle, since the centre can possibly lie anywhere along the line.
A second non-parallel line which also passes the same centre of the circle will now confirm the exact position of the centre of the circle which must be at the intersection of the perpendicular bisector and this second line.
-------------------------------------------------------------
So, our approach is to
1. Find the perpendicular bisector of the particular chord joined by (5, 0) and (3, 4)
--> find midpoint of chord (perpendicular bisector must pass this specific point)
--> find gradient of chord
--> find gradient of perpendicular bisector
--> find the equation of perpendicular bisector
2. Simul our perpendicular bisector with the second equation
3. Solution to the simul is our centre of the circle
4. Use this centre and a point on the circle, namely either (5, 0) or (3, 4), to find our radius
5. Write out the equation of the circle in any form (I prefer standard form because it's more obvious to the eye).
Date Posted:
3 years ago
can the chord be formed using any 2 random points given in the circle? or is there only a way to form chord
im still kind of confused about finding radius of circle tho. the one about using another point on the circle to find radius
im still kind of confused about finding radius of circle tho. the one about using another point on the circle to find radius
Yes, any two points joined together by a line is a chord.
Remember that the radius of a circle is the length between the centre of the circle and any point on the edge/circumference of the circle. The point chosen can be one of the two points used for the chord, because it lies on the circumference.
I have labelled this as the two “r”s in one of the diagrams.
Remember that the radius of a circle is the length between the centre of the circle and any point on the edge/circumference of the circle. The point chosen can be one of the two points used for the chord, because it lies on the circumference.
I have labelled this as the two “r”s in one of the diagrams.
thx for helping me :))
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