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AP EAMCET · Maths · Circle

Find the equation of circle having normals \((x-1)(y-2)=0\) and a tangent \(3 x+4 y=6\) ?

  1. A \((x-1)^2+(y-2)^2=1\)
  2. B \((x-2)^2+(y-1)^2=1\)
  3. C \((x+1)^2+(y+2)^2=1\)
  4. D \((x+2)^2+(y+1)^2=1\)
Verified Solution

Answer & Solution

Correct Answer

(A) \((x-1)^2+(y-2)^2=1\)

Step-by-step Solution

Detailed explanation

The equation of normals to the circle are \(x-1=0\) and \(y-2=0\), so centre of the circle is \((1,2)\) and since \(3 x+4 y=6\) is the tangent to the circle so radius \(r=\frac{3+8-6}{\sqrt{3^2+4^2}}=1\) \(\therefore\) Equation of required circle is \((x-1)^2+(y-2)^2=1\) Hence,…
From AP EAMCET
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