Question:
A circle is such that
\[
(x-2)\cos\theta+(y-2)\sin\theta=1
\]
touches it for all values of \(\theta\). Then the circle is
A circle is such that
\[
(x-2)\cos\theta+(y-2)\sin\theta=1
\]
touches it for all values of \(\theta\). Then the circle is
Show Hint
The equation
\[
(x-h)\cos\theta+(y-k)\sin\theta=r
\]
represents a family of tangents to the circle with centre \((h,k)\) and radius \(r\).
Updated On: Jun 22, 2026
- \(x^2+y^2-4x-4y+7=0\)
- \(x^2+y^2+4x+4y+7=0\)
- \(x^2+y^2-4x-4y-7=0\)
- \(x^2+y^2+4x+4y-7=0\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Identify the family of tangent lines.
Given, \[ (x-2)\cos\theta+(y-2)\sin\theta=1 \] This represents a tangent line to a circle in normal form.
Step 2: Compare with standard tangent form.
The standard tangent form of a circle with centre \((h,k)\) and radius \(r\) is \[ (x-h)\cos\theta+(y-k)\sin\theta=r \] Comparing, \[ h=2,\quad k=2,\quad r=1 \]
Step 3: Write the circle equation.
\[ (x-2)^2+(y-2)^2=1 \]
Step 4: Expand.
\[ x^2-4x+4+y^2-4y+4=1 \] \[ x^2+y^2-4x-4y+7=0 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{x^2+y^2-4x-4y+7=0} \]
Given, \[ (x-2)\cos\theta+(y-2)\sin\theta=1 \] This represents a tangent line to a circle in normal form.
Step 2: Compare with standard tangent form.
The standard tangent form of a circle with centre \((h,k)\) and radius \(r\) is \[ (x-h)\cos\theta+(y-k)\sin\theta=r \] Comparing, \[ h=2,\quad k=2,\quad r=1 \]
Step 3: Write the circle equation.
\[ (x-2)^2+(y-2)^2=1 \]
Step 4: Expand.
\[ x^2-4x+4+y^2-4y+4=1 \] \[ x^2+y^2-4x-4y+7=0 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{x^2+y^2-4x-4y+7=0} \]
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