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# How to work out hard question on a circle?

I found this question on a past paper i'm working on but i don't know how to work it out so can someone please help me. I would really appreciate it :)

A circle has two points P(-1,3) and Q(5,11) as ends of a diameter.

i) Find the equation of the line PQ

ii) Find the co-ordinates of C, the centre of the circle

iii) Find the length of the radius of the circle

iv) Write down the equation of the circle in the form (x-a)^2 + (y-b)^2 = r^2

v) The line y = -7x-4 cuts the circle at P and at another point R. Find coordinates of R

vi) Prove that the line RC is perpendicular to the line PQ

Please can u show all the steps :)

Many many thanks

### 3 Answers

- RogueLv 77 years agoFavorite Answer
m = (y₂ − y₁)/(x₂ − x₁)

=> m = (11 − 3)/(5 − -1)

=> m = 8/6

=> m = 4/3

the using the point-slope formula (y − y₁) = m(x − x₁)

=> y − 3 = (4/3)(x + 1)

=> y − 3 = 4x/3 + 4/3

=> y = 4x/3 + 13/3 <== Answer i

C is the Midpoint of PQ

( (5−1)/2, (11+3)/2 )

=> (2, 7) <== answer ii

d =√( (y₂ − y₁)² + (x₂ − x₁)² ) using C and either P or Q

=> d = √( (11 − 7)² + (5 − 2)² )

=> d = √( 4² + 3² )

=> d = √25

=> d = 5 <== answer iii

(x − 2)² + (y − 7)² = 25 <== answer iv

(x − 2)² + ((-7x − 4) − 7)² = 25

=> (x − 2)² + (-7x − 11)² = 25

=> x² −4x + 4 + 49x² + 154x + 121 = 25

=> 50x² + 150x + 100 = 0

=> 50(x² + 3x + 2) = 0

=> 50(x + 1)(x + 2) = 0

=> x = -1 or -2

as P is x = -1

y = -7(-2) − 4

=> y = 10

R(-2,10) <== answer v

m' = (y₂ − y₁)/(x₂ − x₁) using R and C

=> m' = (7 − 10)/(2 − -2)

=> m' = -3/4

m * m' = -1

=> 4/3 * -3/4 = -1 <== answer vi

- HosamLv 67 years ago
i) Slope of PQ = (11-3)/(5+1) = 8/6=4/3

Hence equation is y - 3 = 4/3(x +1) i.e. y = 4/3 x + 13/3

ii) C = ( 1/2(-1+5), 1/2(3+11) ) = ( 2, 7)

iii) R = 1/2 sqrt( (11-3)^2 + (5 + 1)^2 ) = 1/2 sqrt( 8^2 + 6^2) = 5

iv) The equation is (x - 2)^2 + (y - 7)^2 = 25

v) y = -7x - 4, therefore at the intersection points, (x - 2)^2 + (-7x - 11)^2 = 25

==> (x - 2)^2 + (7x +11)^2 =25

==> x^2 - 4x + 4 + 49 x^2 + 154 x + 121 = 25

==> 50 x^2 + 150 x + 100 = 0

==> x^2 + 3x + 2 = 0

==> (x + 1) (x + 2) = 0

==> x = -1, or x = -2

The corresponding y's are y = -7(-1) - 4 = 3, and y = -7(-2) - 4 = 10

P = (-1, 3), R = (-2, 10)

vi) RC = C - R = (2, 7) - (-2, 10) = (4, -3)

PQ = Q - P = (6, 8)

RC . PQ = 4(6) + (-3)(8) = 0

Hence RC is perpendicular to PQ.

- MechEng2030Lv 77 years ago
i.)

Slope of line PQ = (11 - 3)/(5 + 1) = 8/6 = 4/3

Eqn of line PQ:

y = mx + b where m is the slope

y = 4x/3 + b

Plug either (-1, 3) or (5, 11) in to solve for b:

3 = -4/3 + b => b = 13/3

y = 4x/3 + 13/3

ii.)

C is the midpoint between the points (-1, 3) and (5, 11)

= ((5 - 1)/2, (11 + 3)/2)

= (2, 7)

iii.)

Radius = Distance between center to either P or Q = √((2 + 1)^2 + (4)^2) = 5

iv.)

(a, b) is the center and r is radius, so:

(x - 2)^2 + (y - 7)^2 = 25

v.)

Plugging in y = -7x - 4 into the circle eqn:

(x - 2)^2 + (-7x - 4 - 7)^2 = 25

x^2 - 4x + 4 + 49x^2 + 154x + 121 = 25

50x^2 + 150x + 100 = 0

x^2 + 3x + 2 = 0

(x + 2)(x + 1) = 0

x = -2, -1

So point R is (-2, 10).

vi.)

Slope of line RC = (10 - 7)/(-2 - 2) = -3/4

Slope of line PQ = 4/3

Since slopes of perpendicular lines multiply to -1:

(-3/4)(4/3) = -1

Therefore line RC is perpendicular to line PQ