119638 The equation of the circle passing through the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\) and having centre at \((0,3)\) is
#[Qdiff: Hard, QCat: Numerical Based, examname: Now, So, [JEE Main-2013,COMEDK-2013], As we know that standard equation of ellipse, \(=\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), Comparing of Both (i) and (ii) equation., \(\mathrm{a}^2=16, \mathrm{~b}^2=9\), \(\mathrm{a}=4, \mathrm{~b}=3\), we can see that \(\mathrm{a}>\mathrm{b}\), the foci. of the ellipse will be \(( \pm\) ae, 0 ), We also know that, Eccentricity \(e=\sqrt{1-\frac{b^2}{a^2}}\), Putting value of \(a\) and \(b\) above equation -, \(\mathrm{e}=\sqrt{1-\frac{3^2}{4^2}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}=\frac{\sqrt{7}}{4}\), \(\therefore\) Foci is \(( \pm\) ae, 0\()=\left( \pm 4 \times \frac{\sqrt{7}}{4}, 0\right)=( \pm \sqrt{7}, 0)\), \(\therefore\) Radius of the circle, \(r =\sqrt{(\mathrm{ae})^2+\mathrm{b}^2}\), \(=\sqrt{7+9}=\sqrt{16}=4\), equation of circle of centre \((0,3)\), \((\mathrm{x}-0)^2+(\mathrm{y}-3)^2=16\), \(\therefore \mathrm{x}^2+\mathrm{y}^2-6 \mathrm{y}-7=0\), 17. The circle passing through \((1,-2)\) and touching the axis of \(x\) at \((3,0)\) also passes through the point,

119639 The radius of the circle passing through the origin and making intercepts \(\frac{a}{2}\) and \(\frac{b}{2}\) is

119640 The lines \(2 x-3 y-5=0\) and \(3 x-4 y=7\) are diameters of a circle of area 154 sq units, then the equation of the circle is

119641 The equation \(y^2+3=2(2 x+y)\) represents a parabola with the vertex at
#[Qdiff: Hard, QCat: Numerical Based, examname: And, Therefore, Hence, We known that standard equation of a circle is, \((x-h)^2+(y-k)^2=r^2\), \((x-h)^2+[y-(h-1)]^2=3^2\), \((x-h)^2+(y-h+1)^2=9\), Given that, the circle passes through the point \((7,3)\) and hence we get -, \((7-h)^2+\left[(3-(h-1)]^2=9\right.\), \((7-h)^2+(4-h)^2=9\), \(h^2-11 h+28=0\), \(h^2-7 h-4 h+28=0\), \(h(h-7)-4(h-7)=0\), \((h-7)(h-4)=0\), \(h=4 \text { or h = } 7\), for \(\mathrm{h}=4\) we get \(\mathrm{k}=3\), for \(\mathrm{h}=7\) we get \(\mathrm{k}=6\), the circles which satisfy the given conditions are \((x-7)^2+(y-6)^2=9\), \(\mathrm{x}^2+\mathrm{y}^2-14 \mathrm{x}-12 \mathrm{y}+76=0\), \((x-4)^2+(y-3)^2=9\), \(\text { or, } x^2+y^2-8 x-6 y+16=0\), the required equation of the circle is \(x^2+y^2-\), \(8 x-6 y+16=0\), 22. If \(A\) and \(B\) are two fixed points, then the locus of a point \(P\) which moves in such a way that the \(\angle \mathrm{APB}\) is a right angle, is,

119638 The equation of the circle passing through the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\) and having centre at \((0,3)\) is
#[Qdiff: Hard, QCat: Numerical Based, examname: Now, So, [JEE Main-2013,COMEDK-2013], As we know that standard equation of ellipse, \(=\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), Comparing of Both (i) and (ii) equation., \(\mathrm{a}^2=16, \mathrm{~b}^2=9\), \(\mathrm{a}=4, \mathrm{~b}=3\), we can see that \(\mathrm{a}>\mathrm{b}\), the foci. of the ellipse will be \(( \pm\) ae, 0 ), We also know that, Eccentricity \(e=\sqrt{1-\frac{b^2}{a^2}}\), Putting value of \(a\) and \(b\) above equation -, \(\mathrm{e}=\sqrt{1-\frac{3^2}{4^2}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}=\frac{\sqrt{7}}{4}\), \(\therefore\) Foci is \(( \pm\) ae, 0\()=\left( \pm 4 \times \frac{\sqrt{7}}{4}, 0\right)=( \pm \sqrt{7}, 0)\), \(\therefore\) Radius of the circle, \(r =\sqrt{(\mathrm{ae})^2+\mathrm{b}^2}\), \(=\sqrt{7+9}=\sqrt{16}=4\), equation of circle of centre \((0,3)\), \((\mathrm{x}-0)^2+(\mathrm{y}-3)^2=16\), \(\therefore \mathrm{x}^2+\mathrm{y}^2-6 \mathrm{y}-7=0\), 17. The circle passing through \((1,-2)\) and touching the axis of \(x\) at \((3,0)\) also passes through the point,

119639 The radius of the circle passing through the origin and making intercepts \(\frac{a}{2}\) and \(\frac{b}{2}\) is

119640 The lines \(2 x-3 y-5=0\) and \(3 x-4 y=7\) are diameters of a circle of area 154 sq units, then the equation of the circle is

119641 The equation \(y^2+3=2(2 x+y)\) represents a parabola with the vertex at
#[Qdiff: Hard, QCat: Numerical Based, examname: And, Therefore, Hence, We known that standard equation of a circle is, \((x-h)^2+(y-k)^2=r^2\), \((x-h)^2+[y-(h-1)]^2=3^2\), \((x-h)^2+(y-h+1)^2=9\), Given that, the circle passes through the point \((7,3)\) and hence we get -, \((7-h)^2+\left[(3-(h-1)]^2=9\right.\), \((7-h)^2+(4-h)^2=9\), \(h^2-11 h+28=0\), \(h^2-7 h-4 h+28=0\), \(h(h-7)-4(h-7)=0\), \((h-7)(h-4)=0\), \(h=4 \text { or h = } 7\), for \(\mathrm{h}=4\) we get \(\mathrm{k}=3\), for \(\mathrm{h}=7\) we get \(\mathrm{k}=6\), the circles which satisfy the given conditions are \((x-7)^2+(y-6)^2=9\), \(\mathrm{x}^2+\mathrm{y}^2-14 \mathrm{x}-12 \mathrm{y}+76=0\), \((x-4)^2+(y-3)^2=9\), \(\text { or, } x^2+y^2-8 x-6 y+16=0\), the required equation of the circle is \(x^2+y^2-\), \(8 x-6 y+16=0\), 22. If \(A\) and \(B\) are two fixed points, then the locus of a point \(P\) which moves in such a way that the \(\angle \mathrm{APB}\) is a right angle, is,

119638 The equation of the circle passing through the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\) and having centre at \((0,3)\) is
#[Qdiff: Hard, QCat: Numerical Based, examname: Now, So, [JEE Main-2013,COMEDK-2013], As we know that standard equation of ellipse, \(=\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), Comparing of Both (i) and (ii) equation., \(\mathrm{a}^2=16, \mathrm{~b}^2=9\), \(\mathrm{a}=4, \mathrm{~b}=3\), we can see that \(\mathrm{a}>\mathrm{b}\), the foci. of the ellipse will be \(( \pm\) ae, 0 ), We also know that, Eccentricity \(e=\sqrt{1-\frac{b^2}{a^2}}\), Putting value of \(a\) and \(b\) above equation -, \(\mathrm{e}=\sqrt{1-\frac{3^2}{4^2}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}=\frac{\sqrt{7}}{4}\), \(\therefore\) Foci is \(( \pm\) ae, 0\()=\left( \pm 4 \times \frac{\sqrt{7}}{4}, 0\right)=( \pm \sqrt{7}, 0)\), \(\therefore\) Radius of the circle, \(r =\sqrt{(\mathrm{ae})^2+\mathrm{b}^2}\), \(=\sqrt{7+9}=\sqrt{16}=4\), equation of circle of centre \((0,3)\), \((\mathrm{x}-0)^2+(\mathrm{y}-3)^2=16\), \(\therefore \mathrm{x}^2+\mathrm{y}^2-6 \mathrm{y}-7=0\), 17. The circle passing through \((1,-2)\) and touching the axis of \(x\) at \((3,0)\) also passes through the point,

119639 The radius of the circle passing through the origin and making intercepts \(\frac{a}{2}\) and \(\frac{b}{2}\) is

119640 The lines \(2 x-3 y-5=0\) and \(3 x-4 y=7\) are diameters of a circle of area 154 sq units, then the equation of the circle is

119641 The equation \(y^2+3=2(2 x+y)\) represents a parabola with the vertex at
#[Qdiff: Hard, QCat: Numerical Based, examname: And, Therefore, Hence, We known that standard equation of a circle is, \((x-h)^2+(y-k)^2=r^2\), \((x-h)^2+[y-(h-1)]^2=3^2\), \((x-h)^2+(y-h+1)^2=9\), Given that, the circle passes through the point \((7,3)\) and hence we get -, \((7-h)^2+\left[(3-(h-1)]^2=9\right.\), \((7-h)^2+(4-h)^2=9\), \(h^2-11 h+28=0\), \(h^2-7 h-4 h+28=0\), \(h(h-7)-4(h-7)=0\), \((h-7)(h-4)=0\), \(h=4 \text { or h = } 7\), for \(\mathrm{h}=4\) we get \(\mathrm{k}=3\), for \(\mathrm{h}=7\) we get \(\mathrm{k}=6\), the circles which satisfy the given conditions are \((x-7)^2+(y-6)^2=9\), \(\mathrm{x}^2+\mathrm{y}^2-14 \mathrm{x}-12 \mathrm{y}+76=0\), \((x-4)^2+(y-3)^2=9\), \(\text { or, } x^2+y^2-8 x-6 y+16=0\), the required equation of the circle is \(x^2+y^2-\), \(8 x-6 y+16=0\), 22. If \(A\) and \(B\) are two fixed points, then the locus of a point \(P\) which moves in such a way that the \(\angle \mathrm{APB}\) is a right angle, is,

119638 The equation of the circle passing through the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\) and having centre at \((0,3)\) is
#[Qdiff: Hard, QCat: Numerical Based, examname: Now, So, [JEE Main-2013,COMEDK-2013], As we know that standard equation of ellipse, \(=\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), Comparing of Both (i) and (ii) equation., \(\mathrm{a}^2=16, \mathrm{~b}^2=9\), \(\mathrm{a}=4, \mathrm{~b}=3\), we can see that \(\mathrm{a}>\mathrm{b}\), the foci. of the ellipse will be \(( \pm\) ae, 0 ), We also know that, Eccentricity \(e=\sqrt{1-\frac{b^2}{a^2}}\), Putting value of \(a\) and \(b\) above equation -, \(\mathrm{e}=\sqrt{1-\frac{3^2}{4^2}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}=\frac{\sqrt{7}}{4}\), \(\therefore\) Foci is \(( \pm\) ae, 0\()=\left( \pm 4 \times \frac{\sqrt{7}}{4}, 0\right)=( \pm \sqrt{7}, 0)\), \(\therefore\) Radius of the circle, \(r =\sqrt{(\mathrm{ae})^2+\mathrm{b}^2}\), \(=\sqrt{7+9}=\sqrt{16}=4\), equation of circle of centre \((0,3)\), \((\mathrm{x}-0)^2+(\mathrm{y}-3)^2=16\), \(\therefore \mathrm{x}^2+\mathrm{y}^2-6 \mathrm{y}-7=0\), 17. The circle passing through \((1,-2)\) and touching the axis of \(x\) at \((3,0)\) also passes through the point,

119639 The radius of the circle passing through the origin and making intercepts \(\frac{a}{2}\) and \(\frac{b}{2}\) is

119640 The lines \(2 x-3 y-5=0\) and \(3 x-4 y=7\) are diameters of a circle of area 154 sq units, then the equation of the circle is

119641 The equation \(y^2+3=2(2 x+y)\) represents a parabola with the vertex at
#[Qdiff: Hard, QCat: Numerical Based, examname: And, Therefore, Hence, We known that standard equation of a circle is, \((x-h)^2+(y-k)^2=r^2\), \((x-h)^2+[y-(h-1)]^2=3^2\), \((x-h)^2+(y-h+1)^2=9\), Given that, the circle passes through the point \((7,3)\) and hence we get -, \((7-h)^2+\left[(3-(h-1)]^2=9\right.\), \((7-h)^2+(4-h)^2=9\), \(h^2-11 h+28=0\), \(h^2-7 h-4 h+28=0\), \(h(h-7)-4(h-7)=0\), \((h-7)(h-4)=0\), \(h=4 \text { or h = } 7\), for \(\mathrm{h}=4\) we get \(\mathrm{k}=3\), for \(\mathrm{h}=7\) we get \(\mathrm{k}=6\), the circles which satisfy the given conditions are \((x-7)^2+(y-6)^2=9\), \(\mathrm{x}^2+\mathrm{y}^2-14 \mathrm{x}-12 \mathrm{y}+76=0\), \((x-4)^2+(y-3)^2=9\), \(\text { or, } x^2+y^2-8 x-6 y+16=0\), the required equation of the circle is \(x^2+y^2-\), \(8 x-6 y+16=0\), 22. If \(A\) and \(B\) are two fixed points, then the locus of a point \(P\) which moves in such a way that the \(\angle \mathrm{APB}\) is a right angle, is,