Question 13 of 70intermediate🔧 ApplyLong Answer4 marks

Find the equation of the circle passing through (0,0)(0, 0) and making intercepts aa and bb on the coordinate axes.

Correct Answer

The equation of the required circle is:

x2+y2−ax−by=0x^2 + y^2 - ax - by = 0
Exercise: EXERCISE 10.1 | Q: 13 | (Page 6)
For More Understanding

Explanation

This question tests understanding of circle intercepts and the general equation of a circle. The key insight is that intercepts on coordinate axes give us points on the circle.

An xx-intercept of aa means the point (a,0)(a, 0) lies on the circle, and a yy-intercept of bb means (0,b)(0, b) lies on the circle. Combined with the origin (0,0)(0, 0), we have three points to determine the circle equation.

Solution Steps

  1. Step 1: The circle passes through (0,0)(0, 0) and makes intercepts aa and bb on coordinate axes. This means the circle passes through three points: (0,0)(0, 0), (a,0)(a, 0), and (0,b)(0, b).

  2. Step 2: Let the equation of the circle be:

    x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0

  3. Step 3: Since the circle passes through (0,0)(0, 0), substituting gives:

    c=0c = 0

  4. Step 4: Since the circle passes through (a,0)(a, 0), substituting gives:

    a2+2ga=0a^2 + 2ga = 0

    therefore:

    g=−a2g = -\frac{a}{2}

  5. Step 5: Since the circle passes through (0,b)(0, b), substituting gives:

    b2+2fb=0b^2 + 2fb = 0

    therefore:

    f=−b2f = -\frac{b}{2}

  6. Step 6: Substituting values of gg, ff, and cc in the general equation:

    x2+y2+2(−a2)x+2(−b2)y+0=0x^2 + y^2 + 2\left(-\frac{a}{2}\right)x + 2\left(-\frac{b}{2}\right)y + 0 = 0

  7. Step 7: Simplifying:

    x2+y2−ax−by=0x^2 + y^2 - ax - by = 0
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