div class=trans-pagebuttonPage 1button div class=trans-image amp-img class=trans-thumb alt=Page 1: DSEPP DSE Material · 2018 10 26 · 142 +42 k24 + 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 src=https:reader033fdocumentsusreader033viewer20220528106081bd1545a564362359e138html5thumbnails1jpg width=142 height=106 layout=responsive amp-img divdivdiv class=trans-pagebuttonPage 2button div class=trans-image amp-img class=trans-thumb alt=Page 2: DSEPP DSE Material · 2018 10 26 · 142 +42 k24 + 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 src=https:reader033fdocumentsusreader033viewer20220528106081bd1545a564362359e138html5thumbnails2jpg width=142 height=106 layout=responsive amp-img divdivdiv class=trans-pagebuttonPage 3button div class=trans-image amp-img class=trans-thumb alt=Page 3: DSEPP DSE Material · 2018 10 26 · 142 +42 k24 + 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 src=https:reader033fdocumentsusreader033viewer20220528106081bd1545a564362359e138html5thumbnails3jpg width=142 height=106 layout=responsive amp-img divdivdiv class=trans-pagebuttonPage 4button div class=trans-image amp-img class=trans-thumb alt=Page 4: DSEPP DSE Material · 2018 10 26 · 142 +42 k24 + 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 src=https:reader033fdocumentsusreader033viewer20220528106081bd1545a564362359e138html5thumbnails4jpg width=142 height=106 layout=responsive amp-img divdivdiv class=trans-pagebuttonPage 5button div class=trans-image amp-img class=trans-thumb alt=Page 5: DSEPP DSE Material · 2018 10 26 · 142 +42 k24 + 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...