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 $(x-h)^2+(y-k)^2=a^2$ Here, $A B=8$ units and $L\left(0, \ln \triangle C A M \Rightarrow C A^2=C M^2+A M^2 \Rightarrow\right.$ $C A^2=3^2+4^2 \Rightarrow C A=5 \Rightarrow C L=C A=5 \therefore$ 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 = $5 (x - 5)^2 + (y - 3)^2 = 25, i.e. x^2 + y^2 - 10x - 6y= -9$
Let the required equation be $(x-h)^2+(y-k)^2=a^2$ Here, $A B=8$ units and $L\left(0, \ln \triangle C A M \Rightarrow C A^2=C M^2+A M^2 \Rightarrow\right.$ $C A^2=3^2+4^2 \Rightarrow C A=5 \Rightarrow C L=C A=5 \therefore$ 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 = $5 (x - 5)^2 + (y - 3)^2 = 25, i.e. x^2 + y^2 - 10x - 6y= -9$
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