MHT CET · Maths · Circle
18. The sides of a rectangle are given by the equations \(x=-2, x=4, y=-2\) and \(y=5\)
Then the equation of the circle, whose centre is the point of intersection of the diagonals, lying within the rectangle and touching only two opposite sides, is
- A \(x^2+y^2+2 x+3 y+9=0\)
- B \(x^2+y^2-2 x+3 y+9=0\)
- C \(x^2+y^2+2 x-3 y-9=0\)
- D \(x^2+y^2-2 x-3 y-9=0\)
Answer & Solution
Correct Answer
(D) \(x^2+y^2-2 x-3 y-9=0\)
Step-by-step Solution
Detailed explanation
The given equations of the sides are \(x=-2\), \(x=4, y=-2, y=5\)
\(\therefore \) According to the given condition, centre of the required circle is \(P\).
\(\therefore \) The co-ordinates of \(\mathrm{P}\) are \(\left(1, \frac{3}{2}\right)\).
As circle touches only 2 opposite sides, its radius is either 3.5 units or 3 units.
\(\therefore \) Equation of the required circle is
\(\begin{aligned}
& x^2+y^2-2 x-3 y-\frac{23}{4}=0 \text { or } \\
& x^2+y^2-2 x-3 y-9=0
\end{aligned}\)
\(\therefore \) Option (D) is correct.
\(\therefore \) According to the given condition, centre of the required circle is \(P\).
\(\therefore \) The co-ordinates of \(\mathrm{P}\) are \(\left(1, \frac{3}{2}\right)\).
As circle touches only 2 opposite sides, its radius is either 3.5 units or 3 units.
\(\therefore \) Equation of the required circle is
\(\begin{aligned}
& x^2+y^2-2 x-3 y-\frac{23}{4}=0 \text { or } \\
& x^2+y^2-2 x-3 y-9=0
\end{aligned}\)
\(\therefore \) Option (D) is correct.
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