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Circle C with centre at (-2, 6) passes through the point (10, 11)

Show that the circle C also passes through the point (10,1)

What tripped me up was this question though;
the tangent to the circle C at point (10,11) meets the y axis at point P and the tangent to the circle C at point(10,1) meets y axis at point Q.

How do i show that the distance is 58?
Which distance?
Original post by TheTree0fDeath
Which distance?


Distance PQ
Reply 3
Avatar for mqb2766
mqb2766
21
Not really necessary but the radius is 13, and by symmetry (sketch it), you could just find point P (or Q) and double the distance from the circle y center value (6).

* To find P, the gradient of the line segment to the point (10,11) (from the center) is 5/12, so that gives you the gradient of the tangent as they must be perpendicular (multiply to -1).
* It also therefore gives you the y value of P as you have the gradient of the line and you know it passes through (10,11) so work out the y intercept.
* Subtract 6 and double (or repeat the first to steps to find Q)
Reply 4
Avatar for plopapii
plopapii
2
You need to find the equations of the two tangents and compare their y-intercpets.

Gradient of Line 1
That goes through C and (10,11)
Δy/ Δx

(11-6)/(10-(-2))= 5/12

perpendicular gradient (negative reciprocal) = -12/5

y = (-12/5)x +c

11= (-12/5)10 + c

c=35
Tangent 1:
y= (-12/5)x + 35

Repeat for other line (C and 10,1)
Δy/ Δx

(6-1)/(10-(-2))=5/12

negative reciprocal = -12/5

y=(-12/5)x + c

c= -23

compare y intercepts 35- (-23) = 48
Original post by plopapii
You need to find the equations of the two tangents and compare their y-intercpets.

Gradient of Line 1
That goes through C and (10,11)
Δy/ Δx

(11-6)/(10-(-2))= 5/12

perpendicular gradient (negative reciprocal) = -12/5

y = (-12/5)x +c

11= (-12/5)10 + c

c=35
Tangent 1:
y= (-12/5)x + 35

Repeat for other line (C and 10,1)
Δy/ Δx

(6-1)/(10-(-2))=5/12

negative reciprocal = -12/5

y=(-12/5)x + c

c= -23

compare y intercepts 35- (-23) = 48

Thanks man

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