Find the equations of the circles touching *y*-axis at (0, 3) and making an intercept of 8 units on the *X*-axis.

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#### Solution

**Case I: **The centre lies in first quadrant.

Let the required equation be

\[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]

Here, AB = 8 units and L (0, 3)

In \[\bigtriangleup\]CAM:

In \[\bigtriangleup\]CAM:

\[\Rightarrow C A^2 = C M^2 + A M^2\]

\[\Rightarrow C A^2 = 3^2 + 4^2 \]

\[ \Rightarrow CA = 5\]

\[ \Rightarrow CL = CA = 5\]

\[ \Rightarrow CA = 5\]

\[ \Rightarrow CL = CA = 5\]

∴ Coordinates of the centre = \[\left( 5, 3 \right)\]

And, radius of the circle = 5

\[\left( x - 5 \right)^2 + \left( y - 3 \right)^2 = 25\]

\[x^2 + y^2 - 10x - 6y = - 9\]

**Case II:**The centre lies in the second quadrant.

Coordinates of the centre = \[\left( - 5, 3 \right)\]

And, radius of the circle= 5

\[\left( x + 5 \right)^2 + \left( y - 3 \right)^2 = 25\]

\[x^2 + y^2 + 10x - 6y = - 9\]

Hence, the equation of the required circle is

\[\left( x \pm 5 \right)^2 + \left( y - 3 \right)^2 = 25\]

\[x^2 + y^2 \pm 10x - 6y = - 9\]

Concept: Circle - Standard Equation of a Circle

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