Question
Find the equations of the circles touching y-axis at $(0, 3)$ and making an intercept of $8$ units on the x-axis.
✓
Answer
Case I: The centre lies in first quadrant.
Let the required equation be $(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{a}^2$ Here, $\mathrm{AB}=8$ units and $\mathrm{L}\left(0, \ln \triangle \mathrm{CAM} \Rightarrow \mathrm{CA}^2=\mathrm{CM}^2+\mathrm{AM}^2 \Rightarrow\right.$ $C A^2=3^2+4^2 \Rightarrow C A=5 \Rightarrow C L=C A=5$
$\therefore$ Coordinates of the centre $=(5,3)$ And, radius of the circle $=5(x-5)^2+(y$ $-3)^2=25$, i.e. $x^2+y^2-10 x-6 y=-9$ Case II: The centre lies in the second quadrant.
Coordinates of the centre $=(-5,3)$ And, radius of the circle $=5(x-5)^2+(y-3)^2=25$, i.e. $x^2+y^2-10 x-6 y=-9$
Let the required equation be $(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{a}^2$ Here, $\mathrm{AB}=8$ units and $\mathrm{L}\left(0, \ln \triangle \mathrm{CAM} \Rightarrow \mathrm{CA}^2=\mathrm{CM}^2+\mathrm{AM}^2 \Rightarrow\right.$ $C A^2=3^2+4^2 \Rightarrow C A=5 \Rightarrow C L=C A=5$
$\therefore$ Coordinates of the centre $=(5,3)$ And, radius of the circle $=5(x-5)^2+(y$ $-3)^2=25$, i.e. $x^2+y^2-10 x-6 y=-9$ Case II: The centre lies in the second quadrant.
Coordinates of the centre $=(-5,3)$ And, radius of the circle $=5(x-5)^2+(y-3)^2=25$, i.e. $x^2+y^2-10 x-6 y=-9$
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