Help me with sec 4 circle question amath

miiaow SG

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 :)

miiaow SG

nvm solved.

PhdGuy SG

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

Dalforce 1941 SG

This is sec 4 a math?

PhdGuy SG

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.

ascklee SG

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

PhdGuy SG

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

Hi 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.

wee_ws SG

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

eagle SG

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

Nice.

This gives us the value of a, of which we can find the centre and the radius without the need to do simultaneous equations.

ascklee SG

Wow -- really nice solutions -- both by PhdGuy and by Wen Shih. I learn something everyday. 8-)

Sincerely,

ascklee
OpenlySolved.org

wee_ws SG

Dear eagle and ascklee,

Thanks, let's learn from one another :)

Cheers,
Wen Shih

frekiwang SG

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

PhdGuy SG

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!

PhdGuy SG

Wow -- really nice solutions -- both by PhdGuy and by Wen Shih. I learn something everyday. 8-)

Sincerely,

ascklee
OpenlySolved.org

me too. I am learning something new everyday.