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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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प्रश्न

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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उत्तर

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:
\[\Rightarrow C A^2 = C M^2 + A M^2\]
\[\Rightarrow C A^2 = 3^2 + 4^2 \]
\[ \Rightarrow CA = 5\]
\[ \Rightarrow CL = CA = 5\]
∴ 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
\[\left( x \pm 5 \right)^2 + \left( y - 3 \right)^2 = 25\]
\[x^2 + y^2 \pm 10x - 6y = - 9\]
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Advanced Concept of Circle - Standard Equation of a Circle
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अध्याय 24: The circle - Exercise 24.1 [पृष्ठ २१]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 24 The circle
Exercise 24.1 | Q 12 | पृष्ठ २१

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