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

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

119641 The equation $y^2+3=2(2x+y)$ 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 -, $(7-h{)}^2+[(3-(h-1){]}^2=9$, $(7-h{)}^2+(4-h{)}^2=9$, $h^2-11h+28=0$, $h^2-7h-4h+28=0$, $h(h-7)-4(h-7)=0$, $(h-7)(h-4)=0$, $h=4\text{ or h = }7$, 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 $(x-7{)}^2+(y-6{)}^2=9$, ${\mathrm{x}}^2+{\mathrm{y}}^2-14\mathrm{x}-12\mathrm{y}+76=0$, $(x-4{)}^2+(y-3{)}^2=9$, $\text{ or, }{x}^2+y^2-8x-6y+16=0$, the required equation of the circle is $x^2+y^2-$, $8x-6y+16=0$, 22. If $A$ and $B$ are two fixed points, then the locus of a point $P$ which moves in such a way that the $\mathrm{\angle }\mathrm{APB}$ 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$)=(\pm 4\times \frac{\sqrt{7}}{4},0)=(\pm \sqrt{7},0)$, $\therefore$ 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 $2x-3y-5=0$ and $3x-4y=7$ are diameters of a circle of area 154 sq units, then the equation of the circle is

119641 The equation $y^2+3=2(2x+y)$ 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 -, $(7-h{)}^2+[(3-(h-1){]}^2=9$, $(7-h{)}^2+(4-h{)}^2=9$, $h^2-11h+28=0$, $h^2-7h-4h+28=0$, $h(h-7)-4(h-7)=0$, $(h-7)(h-4)=0$, $h=4\text{ or h = }7$, 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 $(x-7{)}^2+(y-6{)}^2=9$, ${\mathrm{x}}^2+{\mathrm{y}}^2-14\mathrm{x}-12\mathrm{y}+76=0$, $(x-4{)}^2+(y-3{)}^2=9$, $\text{ or, }{x}^2+y^2-8x-6y+16=0$, the required equation of the circle is $x^2+y^2-$, $8x-6y+16=0$, 22. If $A$ and $B$ are two fixed points, then the locus of a point $P$ which moves in such a way that the $\mathrm{\angle }\mathrm{APB}$ 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$)=(\pm 4\times \frac{\sqrt{7}}{4},0)=(\pm \sqrt{7},0)$, $\therefore$ 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 $2x-3y-5=0$ and $3x-4y=7$ are diameters of a circle of area 154 sq units, then the equation of the circle is

119641 The equation $y^2+3=2(2x+y)$ 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 -, $(7-h{)}^2+[(3-(h-1){]}^2=9$, $(7-h{)}^2+(4-h{)}^2=9$, $h^2-11h+28=0$, $h^2-7h-4h+28=0$, $h(h-7)-4(h-7)=0$, $(h-7)(h-4)=0$, $h=4\text{ or h = }7$, 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 $(x-7{)}^2+(y-6{)}^2=9$, ${\mathrm{x}}^2+{\mathrm{y}}^2-14\mathrm{x}-12\mathrm{y}+76=0$, $(x-4{)}^2+(y-3{)}^2=9$, $\text{ or, }{x}^2+y^2-8x-6y+16=0$, the required equation of the circle is $x^2+y^2-$, $8x-6y+16=0$, 22. If $A$ and $B$ are two fixed points, then the locus of a point $P$ which moves in such a way that the $\mathrm{\angle }\mathrm{APB}$ 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$)=(\pm 4\times \frac{\sqrt{7}}{4},0)=(\pm \sqrt{7},0)$, $\therefore$ 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,