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samnatha

please help find the equations of the tangents to these circles at the points given x^2 + y^2 +2x + 4y - 12 = 0 (3, -1)

  • one year ago
  • one year ago

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  1. campbell_st
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    the circle will have a tangent at x = 3 which is in the top half of the circle as the centre is (-1, -2) you need to get the equation of the circle with y as the subject (x + 1)^2 + (y + 2)^2 = 12 - 1 - 4 or (x +1)^2 + (y + 2)^2 = 7 (y + 2)^2 = 7 - (x + 1)^2 \[y + 2 = \pm \sqrt{7 - (x + 1)^2}\] so \[y = \pm \sqrt{7 - (x + 1)^2} - 2\] so you need to differentiate the which is the upper semicircle \[y = \sqrt{7 - (x + 1)^2} - 2\] and then substitute the find the slope of the tangent... once you have the slope... then use the point (3, -1) to find the equations... hope it helps.

    • one year ago
  2. samnatha
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    i cam't differentiate as we haven't learnt it yet ?

    • one year ago
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