119723 A circle with centre at \((3,6)\) passes through \((-1\),
1). Its equation is
#[Qdiff: Hard, QCat: Numerical Based, examname: Kerala CEE-2013], So, We know that, standard equation of a circle is -, \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\), \((\mathrm{x}-2)^2+(\mathrm{y}-2)^2=\mathrm{r}^2\), \(\because\) Centre passes through \((4,5)\), \((4-2)^2+(5-2)^2=r^2\), \(4+9=r^2\), \(13=r^2\), the equation of circle is -, \((x-2)^2+(y-2)^2=13\), \(x^2+4-4 x+y^2+4-4 y=13\), \(x^2+y^2-4 x-4 y-5=0\), 112. The equation of the circle which touches the lines \(x=0, y=0\) and \(4 x+3 y=12\) is,

119724 The parametric equation of the circle \(\mathbf{x}^2+\mathbf{y}^2+\) \(x+\sqrt{3} y=0\) are

119725 The line segment joining the points \((4,7)\) and \((-2,-1)\) is a diameter of a circle. If the circle intersects the \(x\)-axis at \(A\) and \(B\), then \(A B\) is equal to

119651 The equation of a circle passing through origin and radius is \(a\), is

119723 A circle with centre at \((3,6)\) passes through \((-1\),
1). Its equation is
#[Qdiff: Hard, QCat: Numerical Based, examname: Kerala CEE-2013], So, We know that, standard equation of a circle is -, \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\), \((\mathrm{x}-2)^2+(\mathrm{y}-2)^2=\mathrm{r}^2\), \(\because\) Centre passes through \((4,5)\), \((4-2)^2+(5-2)^2=r^2\), \(4+9=r^2\), \(13=r^2\), the equation of circle is -, \((x-2)^2+(y-2)^2=13\), \(x^2+4-4 x+y^2+4-4 y=13\), \(x^2+y^2-4 x-4 y-5=0\), 112. The equation of the circle which touches the lines \(x=0, y=0\) and \(4 x+3 y=12\) is,

119724 The parametric equation of the circle \(\mathbf{x}^2+\mathbf{y}^2+\) \(x+\sqrt{3} y=0\) are

119725 The line segment joining the points \((4,7)\) and \((-2,-1)\) is a diameter of a circle. If the circle intersects the \(x\)-axis at \(A\) and \(B\), then \(A B\) is equal to

119651 The equation of a circle passing through origin and radius is \(a\), is

119723 A circle with centre at \((3,6)\) passes through \((-1\),
1). Its equation is
#[Qdiff: Hard, QCat: Numerical Based, examname: Kerala CEE-2013], So, We know that, standard equation of a circle is -, \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\), \((\mathrm{x}-2)^2+(\mathrm{y}-2)^2=\mathrm{r}^2\), \(\because\) Centre passes through \((4,5)\), \((4-2)^2+(5-2)^2=r^2\), \(4+9=r^2\), \(13=r^2\), the equation of circle is -, \((x-2)^2+(y-2)^2=13\), \(x^2+4-4 x+y^2+4-4 y=13\), \(x^2+y^2-4 x-4 y-5=0\), 112. The equation of the circle which touches the lines \(x=0, y=0\) and \(4 x+3 y=12\) is,

119724 The parametric equation of the circle \(\mathbf{x}^2+\mathbf{y}^2+\) \(x+\sqrt{3} y=0\) are

119725 The line segment joining the points \((4,7)\) and \((-2,-1)\) is a diameter of a circle. If the circle intersects the \(x\)-axis at \(A\) and \(B\), then \(A B\) is equal to

119651 The equation of a circle passing through origin and radius is \(a\), is

119723 A circle with centre at \((3,6)\) passes through \((-1\),
1). Its equation is
#[Qdiff: Hard, QCat: Numerical Based, examname: Kerala CEE-2013], So, We know that, standard equation of a circle is -, \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\), \((\mathrm{x}-2)^2+(\mathrm{y}-2)^2=\mathrm{r}^2\), \(\because\) Centre passes through \((4,5)\), \((4-2)^2+(5-2)^2=r^2\), \(4+9=r^2\), \(13=r^2\), the equation of circle is -, \((x-2)^2+(y-2)^2=13\), \(x^2+4-4 x+y^2+4-4 y=13\), \(x^2+y^2-4 x-4 y-5=0\), 112. The equation of the circle which touches the lines \(x=0, y=0\) and \(4 x+3 y=12\) is,

119724 The parametric equation of the circle \(\mathbf{x}^2+\mathbf{y}^2+\) \(x+\sqrt{3} y=0\) are

119725 The line segment joining the points \((4,7)\) and \((-2,-1)\) is a diameter of a circle. If the circle intersects the \(x\)-axis at \(A\) and \(B\), then \(A B\) is equal to

119651 The equation of a circle passing through origin and radius is \(a\), is