Help me with sec 4 circle question amath
-
Find the equation of the circle which passes thru the points *(-6,5), Q(2,1) and has its centre lying on the line y=x+4.
thanks :)
-
nvm solved.
-
Ans: center = (-3,1)
Radius = 5
Solution:
equation of circle: (x-a)^2 + (y-b)^2 = r^2
since center lies on y=x+4, => b = a+4
subst into equation : (x-a)^2 + [y-(a+4)]^2 = r^2
since it satisfies two points:
eq (1): (-6-a)^2 + [5-(a+4)]^2 = r^2
eq (2): (2-a)^2 + [1-(a+4)]^2 = r^2
(1) - (2) to eliminate r: NOTE: make use of identity: a^2-b^2 = (a+b)(a-b)
Solve for a = -3
then b=a+4 = 1
subst a & b into equation of circle to get r = 5
-
This is sec 4 a math?
-
just for the fun of it, let me modify this to become a Physics question:
A small bead with a mass of 0.2kg rolls along the chord starting from rest at point P(-6,5) to point Q(2,1). It leaves the chord at Q(2,1) and continues its motion under the force of gravity. Assuming the chord is frictionless and ignoring rolling inertia & air resistance, what is its velocity when it crosses the x-axis? Will the mass of the ball affect the answer? Assume g = -9.81m/s^2.
-
Yes, this is Sec 4 maths.
Questions similar to this are:
[1] Panpac Additional Mathematics - Exercise 9.2 - Question 6 (p. 194)
[2] Panpac Additional Mathematics - Miscellaneous Exercise 9 - Question 8 (p. 198)
Sincerely,
ascklee
OpenlySolved.org -
Originally posted by ascklee:
Yes, this is Sec 4 maths.
Questions similar to this are:
[1] Panpac Additional Mathematics - Exercise 9.2 - Question 6 (p. 194)
[2] Panpac Additional Mathematics - Miscellaneous Exercise 9 - Question 8 (p. 198)
Sincerely,
ascklee
OpenlySolved.orgHi Dr ascklee,
Do you have a more elegant way to solve the above problem. I think my solution is a bit 'contrived'. I am always looking for elegant ways to solve elementary problems.
thanks.
-
Hi,
Let the centre be (a, a + 4).
Consider the fact that
distance between (-6, 5) and (a, a + 4) = distance between (2, 1) and (a, a + 4),
from which we square both sides and solve for the value of a.
Thanks.
Cheers,
Wen Shih -
Originally posted by wee_ws:
Hi,
Let the centre be (a, a + 4).
Consider the fact that
distance between (-6, 5) and (a, a + 4) = distance between (2, 1) and (a, a + 4),
from which we square both sides and solve for the value of a.
Thanks.
Cheers,
Wen ShihNice.
This gives us the value of a, of which we can find the centre and the radius without the need to do simultaneous equations.
-
Wow -- really nice solutions -- both by PhdGuy and by Wen Shih. I learn something everyday. 8-)
Sincerely,
ascklee
OpenlySolved.org -
Dear eagle and ascklee,
Thanks, let's learn from one another :)
Cheers,
Wen Shih -
the centre must lie on the perpendicular bisector of PQ,
mid-point of PQ is (-2,3), gradient of PQ = -0.5
gradient of the perpendicular bisector = 2, equation of the p.b. y-3=2(x+2) (i.e. y=2x+7)
since the centre is also given to lie on y=x+4
solve simutaneously, we have x=-3, y=1
so centre at (-3,1), radius^2=(-3-2)^2+(1-1)^2=25 (centre to Q)
equation of circle: (x+3)^2+(y-1)^2=25
-
Originally posted by frekiwang:
the centre must lie on the perpendicular bisector of PQ,
mid-point of PQ is (-2,3), gradient of PQ = -0.5
gradient of the perpendicular bisector = 2, equation of the p.b. y-3=2(x+2) (i.e. y=2x+7)
since the centre is also given to lie on y=x+4
solve simutaneously, we have x=-3, y=1
so centre at (-3,1), radius^2=(-3-2)^2+(1-1)^2=25 (centre to Q)
equation of circle: (x+3)^2+(y-1)^2=25
bingo!. This is the alternative solution I was looking for!
-
Originally posted by ascklee:
Wow -- really nice solutions -- both by PhdGuy and by Wen Shih. I learn something everyday. 8-)
Sincerely,
ascklee
OpenlySolved.orgme too. I am learning something new everyday.
