119717 Two points from the set of concyclic points of the circle passing through \((1,1),(2,-1),(3,2)\) is

119718 If the equation \(\lambda x^2+(2 \lambda-3) y^2-4 x-1=0\) represents a circle, then its radius is

119719 If the \(y=7 x-25\) meets the circle \(x^2+y^2=25\) in the points A, B then the distance between \(A\) and \(B\) is

119720 The equation of the circle is \(3 x^2+3 y^2+6 x-4 y\) \(-1=0\). Then its radius is
#[Qdiff: Hard, QCat: Numerical Based, examname: And, [Kerala CEE-2022], Compare with standard equation which is, From equation (i), Here, \(\quad 2 \mathrm{~g}=2\), \(\Rightarrow \quad \mathrm{g}=1\), \(\mathrm{c}=\frac{-1}{3}\), Then radius, \(\quad \mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\), \(r=\sqrt{1^2+\left(\frac{-2}{3}\right)^2-\left(\frac{-1}{3}\right)}\), \(r=\sqrt{1+\frac{4}{9}+\frac{1}{3}}\), \(r=\sqrt{\frac{9+4+3}{9}}\), \(r=\sqrt{\frac{16}{9}}\), \(r=\frac{4}{3}\), 107. If the radius of the circle \(x^2+y^2+a x+b y+3=\) 0 is 2 , then the point \((a, b)\) lies on the circle,

119721 The equation of the circle whose radius is \(\sqrt{7}\) and concentric with the circle \(x^2+y^2-8 x+6 y\) \(-11=0\) is

119717 Two points from the set of concyclic points of the circle passing through \((1,1),(2,-1),(3,2)\) is

119718 If the equation \(\lambda x^2+(2 \lambda-3) y^2-4 x-1=0\) represents a circle, then its radius is

119719 If the \(y=7 x-25\) meets the circle \(x^2+y^2=25\) in the points A, B then the distance between \(A\) and \(B\) is

119720 The equation of the circle is \(3 x^2+3 y^2+6 x-4 y\) \(-1=0\). Then its radius is
#[Qdiff: Hard, QCat: Numerical Based, examname: And, [Kerala CEE-2022], Compare with standard equation which is, From equation (i), Here, \(\quad 2 \mathrm{~g}=2\), \(\Rightarrow \quad \mathrm{g}=1\), \(\mathrm{c}=\frac{-1}{3}\), Then radius, \(\quad \mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\), \(r=\sqrt{1^2+\left(\frac{-2}{3}\right)^2-\left(\frac{-1}{3}\right)}\), \(r=\sqrt{1+\frac{4}{9}+\frac{1}{3}}\), \(r=\sqrt{\frac{9+4+3}{9}}\), \(r=\sqrt{\frac{16}{9}}\), \(r=\frac{4}{3}\), 107. If the radius of the circle \(x^2+y^2+a x+b y+3=\) 0 is 2 , then the point \((a, b)\) lies on the circle,

119721 The equation of the circle whose radius is \(\sqrt{7}\) and concentric with the circle \(x^2+y^2-8 x+6 y\) \(-11=0\) is

119717 Two points from the set of concyclic points of the circle passing through \((1,1),(2,-1),(3,2)\) is

119718 If the equation \(\lambda x^2+(2 \lambda-3) y^2-4 x-1=0\) represents a circle, then its radius is

119719 If the \(y=7 x-25\) meets the circle \(x^2+y^2=25\) in the points A, B then the distance between \(A\) and \(B\) is

119720 The equation of the circle is \(3 x^2+3 y^2+6 x-4 y\) \(-1=0\). Then its radius is
#[Qdiff: Hard, QCat: Numerical Based, examname: And, [Kerala CEE-2022], Compare with standard equation which is, From equation (i), Here, \(\quad 2 \mathrm{~g}=2\), \(\Rightarrow \quad \mathrm{g}=1\), \(\mathrm{c}=\frac{-1}{3}\), Then radius, \(\quad \mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\), \(r=\sqrt{1^2+\left(\frac{-2}{3}\right)^2-\left(\frac{-1}{3}\right)}\), \(r=\sqrt{1+\frac{4}{9}+\frac{1}{3}}\), \(r=\sqrt{\frac{9+4+3}{9}}\), \(r=\sqrt{\frac{16}{9}}\), \(r=\frac{4}{3}\), 107. If the radius of the circle \(x^2+y^2+a x+b y+3=\) 0 is 2 , then the point \((a, b)\) lies on the circle,

119721 The equation of the circle whose radius is \(\sqrt{7}\) and concentric with the circle \(x^2+y^2-8 x+6 y\) \(-11=0\) is

119717 Two points from the set of concyclic points of the circle passing through \((1,1),(2,-1),(3,2)\) is

119718 If the equation \(\lambda x^2+(2 \lambda-3) y^2-4 x-1=0\) represents a circle, then its radius is

119719 If the \(y=7 x-25\) meets the circle \(x^2+y^2=25\) in the points A, B then the distance between \(A\) and \(B\) is

119720 The equation of the circle is \(3 x^2+3 y^2+6 x-4 y\) \(-1=0\). Then its radius is
#[Qdiff: Hard, QCat: Numerical Based, examname: And, [Kerala CEE-2022], Compare with standard equation which is, From equation (i), Here, \(\quad 2 \mathrm{~g}=2\), \(\Rightarrow \quad \mathrm{g}=1\), \(\mathrm{c}=\frac{-1}{3}\), Then radius, \(\quad \mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\), \(r=\sqrt{1^2+\left(\frac{-2}{3}\right)^2-\left(\frac{-1}{3}\right)}\), \(r=\sqrt{1+\frac{4}{9}+\frac{1}{3}}\), \(r=\sqrt{\frac{9+4+3}{9}}\), \(r=\sqrt{\frac{16}{9}}\), \(r=\frac{4}{3}\), 107. If the radius of the circle \(x^2+y^2+a x+b y+3=\) 0 is 2 , then the point \((a, b)\) lies on the circle,

119721 The equation of the circle whose radius is \(\sqrt{7}\) and concentric with the circle \(x^2+y^2-8 x+6 y\) \(-11=0\) is

119717 Two points from the set of concyclic points of the circle passing through \((1,1),(2,-1),(3,2)\) is

119718 If the equation \(\lambda x^2+(2 \lambda-3) y^2-4 x-1=0\) represents a circle, then its radius is

119719 If the \(y=7 x-25\) meets the circle \(x^2+y^2=25\) in the points A, B then the distance between \(A\) and \(B\) is

119720 The equation of the circle is \(3 x^2+3 y^2+6 x-4 y\) \(-1=0\). Then its radius is
#[Qdiff: Hard, QCat: Numerical Based, examname: And, [Kerala CEE-2022], Compare with standard equation which is, From equation (i), Here, \(\quad 2 \mathrm{~g}=2\), \(\Rightarrow \quad \mathrm{g}=1\), \(\mathrm{c}=\frac{-1}{3}\), Then radius, \(\quad \mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\), \(r=\sqrt{1^2+\left(\frac{-2}{3}\right)^2-\left(\frac{-1}{3}\right)}\), \(r=\sqrt{1+\frac{4}{9}+\frac{1}{3}}\), \(r=\sqrt{\frac{9+4+3}{9}}\), \(r=\sqrt{\frac{16}{9}}\), \(r=\frac{4}{3}\), 107. If the radius of the circle \(x^2+y^2+a x+b y+3=\) 0 is 2 , then the point \((a, b)\) lies on the circle,

119721 The equation of the circle whose radius is \(\sqrt{7}\) and concentric with the circle \(x^2+y^2-8 x+6 y\) \(-11=0\) is