Question
Download Solution PDFFind the equation of the circle that passes through the points (-1, 0), (1, 0) and (0, -1).
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The general form of the equation of a circle is:
x2 + y2 + 2gx + 2fy + c = 0
- The center of the circle is (-g, -f).
-
The radius of the circle is \(\sqrt{g^{2}+f^{2}-c}\)
Calculation:
Let the required circle be x2 + y2 + 2gx + 2fy + c = 0 ----(1)
It passes through (-1, 0), (1, 0) and (0, -1).
Therefore, on substituting the coordinates of three points successively in equation (1), we get
1 - 2g + c = 0 ----(2)
1 + 2g + c = 0 ----(3)
1 - 2f + c = 0 ----(4)
Subtracting equation (2) from the (3), we get
4g = 0
⇒ g = 0
On putting the value of g in equation (3), we get
c = -1
Now, putting the value of c in equation (4), we get
1 - 2f - 1 = 0
⇒ f = 0
On substituting the value of g, f, and c in equation (1), we get
x2 + y2 - 1 = 0
⇒ x2 + y2 = 1
Hence, the required equation is x2 + y2 = 1.
Last updated on May 31, 2025
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