Question:
For a circle of diameter \(R\), touching
\[
x^2+y^2-4y=0
\]
and passing through \((4,5)\), which of the following is correct?
For a circle of diameter \(R\), touching
\[
x^2+y^2-4y=0
\]
and passing through \((4,5)\), which of the following is correct?
Show Hint
If two circles touch each other, the distance between their centres equals either the sum or the difference of their radii depending on external or internal touching.
Updated On: Jun 24, 2026
- \(3\leq R\leq 7\)
- \(0\lt R\lt 3\)
- \(R\gt 7\)
- \(\frac{3}{2}\leq R\leq \frac{7}{2}\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the given circle.
The equation \[ x^2+y^2-4y=0 \] can be written as \[ x^2+(y-2)^2=4 \] Thus, the given circle has centre \[ (0,2) \] and radius \[ 2 \]
Step 2: Use the condition of touching circles.
Suppose the required circle has centre \((h,k)\) and radius \(r\).
Since its diameter is \(R\), \[ R=2r \] The circle passes through \((4,5)\), so \[ \sqrt{(h-4)^2+(k-5)^2}=r \] Also, since it touches the given circle externally or internally, \[ \sqrt{h^2+(k-2)^2}=r+2 \] or \[ \sqrt{h^2+(k-2)^2}=|r-2| \]
Step 3: Use geometric interpretation.
The point \((4,5)\) lies at distance \[ \sqrt{4^2+(5-2)^2} = \sqrt{16+9} = 5 \] from the centre \((0,2)\) of the given circle.
For touching circles passing through this point, the possible radius values satisfy \[ \frac{5-2}{2}\leq r\leq \frac{5+2}{2} \] \[ \frac{3}{2}\leq r\leq \frac{7}{2} \] Since \[ R=2r, \] multiplying throughout by \(2\), \[ 3\leq R\leq 7 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{3\leq R\leq 7} \]
The equation \[ x^2+y^2-4y=0 \] can be written as \[ x^2+(y-2)^2=4 \] Thus, the given circle has centre \[ (0,2) \] and radius \[ 2 \]
Step 2: Use the condition of touching circles.
Suppose the required circle has centre \((h,k)\) and radius \(r\).
Since its diameter is \(R\), \[ R=2r \] The circle passes through \((4,5)\), so \[ \sqrt{(h-4)^2+(k-5)^2}=r \] Also, since it touches the given circle externally or internally, \[ \sqrt{h^2+(k-2)^2}=r+2 \] or \[ \sqrt{h^2+(k-2)^2}=|r-2| \]
Step 3: Use geometric interpretation.
The point \((4,5)\) lies at distance \[ \sqrt{4^2+(5-2)^2} = \sqrt{16+9} = 5 \] from the centre \((0,2)\) of the given circle.
For touching circles passing through this point, the possible radius values satisfy \[ \frac{5-2}{2}\leq r\leq \frac{5+2}{2} \] \[ \frac{3}{2}\leq r\leq \frac{7}{2} \] Since \[ R=2r, \] multiplying throughout by \(2\), \[ 3\leq R\leq 7 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{3\leq R\leq 7} \]
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