Advertisement Remove all ads

Find the Equation of a Circle Which Touches Both the Axes and Passes Through the Point (2, 1). - Mathematics

Find the equation of a circle
which touches both the axes and passes through the point (2, 1).

Advertisement Remove all ads

Solution

Let (hk) be the centre of a circle with radius a.
Thus, its equation will be \[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]

Let the required equation of the circle be

\[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]

It is given that the circle touches both the axes.
Thus, the required equation will be

\[x^2 + y^2 - 2ax - 2ay + a^2 = 0\]

Also, the circle passes through the point (2, 1).
∴ \[4 + 1 - 4a - 2a + a^2 = 0\]

\[\Rightarrow a^2 - 6a + 5 = 0\]
\[ \Rightarrow a^2 - 5a - a + 5 = 0\]
\[ \Rightarrow a = 1, 5\]

Hence, the required equation is \[x^2 + y^2 - 2x - 2y + 1 = 0\] or

\[x^2 + y^2 - 10x - 10y + 25 = 0\]
Concept: Circle - Standard Equation of a Circle
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 24 The circle
Exercise 24.1 | Q 7.3 | Page 21
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×