Question

If the radius of the circle x²+y²+ax+by+3=0 is 2, then the point (a, b) lies on the circle

• A. x²+y²=7
• B. x²+y²=4
• C. x²+y²=14
• D. x²+y²=28
• E. x²+y²=1

Answer 1

The Correct Option is D

The general equation of a circle is \( x^2 + y^2 + ax + by + c = 0 \), where \( (h, k) \) is the center and \( r \) is the radius. We are given the equation of the circle as \( x^2 + y^2 + ax + by + 3 = 0 \). The radius of the circle is given as 2, so we use the formula for the radius of a circle: \[ r = \sqrt{h^2 + k^2 - c}, \] where \( h = -\frac{a}{2} \), \( k = -\frac{b}{2} \), and \( c = 3 \). Substitute the values for \( h \) and \( k \) into the formula and simplify: \[ r = \sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2 - 3}. \] We know that the radius is 2, so we have: \[ 2 = \sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2 - 3}. \] Squaring both sides: \[ 4 = \left( \frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2 - 3. \] Simplifying: \[ 7 = \left( \frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2. \] Thus, the point \( (a, b) \) lies on the circle given by the equation \( x^2 + y^2 = 28 \).

The correct option is (D) : \(x^2+y^2=28\)

Answer 2

The general equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\), where the center is \((-g, -f)\) and the radius is \(r = \sqrt{g^2 + f^2 - c}\).

Comparing the given equation \(x^2 + y^2 + ax + by + 3 = 0\) with the general equation, we have \(2g = a\), \(2f = b\), and \(c = 3\). So, \(g = \frac{a}{2}\) and \(f = \frac{b}{2}\).

The radius is given as 2. Therefore, \(r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 - 3} = 2\).

Squaring both sides, we get:

\(\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 - 3 = 4\)

Multiplying by 4, we have:

\(a^2 + b^2 - 12 = 16\)

Adding 12 to both sides, we get:

\(a^2 + b^2 = 28\)

Since we want to find the equation of the circle that the point (a, b) lies on, we replace a with x and b with y, which gives us:

\(x^2 + y^2 = 28\)

Therefore, the point (a, b) lies on the circle \(x^2 + y^2 = 28\).