DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 =...

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Transcript of DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 =...

Page 1: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 2: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 3: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 4: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 5: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 6: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 7: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 8: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 9: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 10: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 11: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 12: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the
Page 13: DSEPP | DSE Material · 2018. 10. 26. · 142 +42 k2(4) + k 3 = 0 and k3 = 92 = o 82 +122 42 +42 = 0 Solving, we have kl = -18, So, the equation of the circle is x The slope of the