Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

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

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  • Rogue
    Lv 7
    7 years ago
    Favorite 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

  • Hosam
    Lv 6
    7 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.

  • 7 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

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