119786
If one of the diameters of the circle, given by the equation, \(x^2+y^2-4 x+6 y-12=0\), is a chord of a circle \(S\), whose centre is at \((-3,2)\) then the radius of \(S\) is
1 \(5 \sqrt{2}\)
2 \(5 \sqrt{3}\)
3 5
4 10
Explanation:
B Given circle equation- \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}-12=0\) \(\text { From comparing general equation of circle we get }\) \(\quad \mathrm{g}_1=+2, \mathrm{f},=-3 \quad \mathrm{c}_1=-12\) \(\mathrm{r}_1=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}_1}\) \(\mathrm{r}_1=\sqrt{4+9+12}=5\) \(\mathrm{C}_2=(-3,2)\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(5)^2+(5)^2}\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{50}\) \(\mathrm{C}_2 \mathrm{~A}=\sqrt{(5)^2+(\sqrt{50})^2}=\sqrt{75}\) \(\mathrm{C}_2 \mathrm{~A}=\mathrm{r}=5 \sqrt{3}\)
JEE Main-2016
Conic Section
119787
A circle cuts a chord of length 4 a on the \(\mathrm{X}\)-axis and passes through a point on the \(\mathrm{Y}\)-axis, distant \(2 \mathrm{~b}\) from the origin. Then the locus of the centre of this circle, is
1 a parabola
2 an ellipse
3 a straight line
4 a hyperbola
Explanation:
A Let equation of the circle is, \(x^2+y^2+2 g x+2 f y+c=0\) Center \(=(-\mathrm{g},-\mathrm{f})\) Radius \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) The circle equation passes through \((0,2 b)\) \(4 \mathrm{~b}^2+4 \mathrm{fb}+\mathrm{c}=0\) intercept on \(\mathrm{x}-\) axis \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}=4 \mathrm{a}\) \(\mathrm{g}^2-\mathrm{c}=4 \mathrm{a}^2\) \(\mathrm{c}=\mathrm{g}^2-4 \mathrm{a}^2\) Putting in eq - (i) \(4 b^2+4 f+g^2-4 a^2=0\) \(x^2+4 y+4\left(b^2-a^2\right)=0\)Its represent parabola.
JEE Main 11.01.2019
Conic Section
119788
A square is inscribed in the circle \(x^2+y^2-6 x+\) \(8 y-103=0\) with its sides parallel to the coordinate axes. Then, the distance of the vertex of this square which is nearest to the origin is
1 6
2 13
3 \(\sqrt{41}\)
4 \(\sqrt{137}\)
Explanation:
C Given, \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+8 \mathrm{y}-103=0\) \((-\mathrm{g},-\mathrm{f})=3,-4\) \(\mathrm{c}=-103\) \(\text { From figure, }\) \(\mathrm{R}=\sqrt{9+16+103}=8 \sqrt{2}\) \(\mathrm{OB}=\sqrt{265}, \mathrm{OC}=\sqrt{137}, \mathrm{OD}=\sqrt{41}\) \(\text { Hence OD }=\sqrt{41} \text { is the nearest to the origin }\) From figure, Hence OD \(=\sqrt{41}\) is the nearest to the origin
JEE Main 11.01.2019
Conic Section
119789
A rectangle is inscribed in a circle with a diameter lying along the line \(3 y=x+7\). If the two adjacent vertices of the rectangle are \((-8\), 5 ) and \((6,5)\), then the area of the rectangle (in sq units) is
1 72
2 84
3 98
4 56
Explanation:
B From figure, \(\text { Let vertex } \mathrm{C} \text { is }(6, \mathrm{k})\) \(\therefore \text { centre of circle is }\) \(\left(\frac{-8+6}{2}, \frac{5+\mathrm{k}}{2}\right)=\left(-1, \frac{5+\mathrm{k}}{2}\right)\) Which is lying on diameter of circle \(3 \mathrm{y}=\mathrm{x}+7\) \(3\left(\frac{5+\mathrm{k}}{2}\right)=-1+7\) \(15+3 \mathrm{k}=12\) \(\mathrm{k}=-1\) \(\text { vertex of } \mathrm{C}=(6,-1)\) \(\text { Length of } A B=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(=\sqrt{(6+8)^2+(5-5)^2}=14\) \(\text { Length of } B C=\sqrt{(6-6)^2+(-1-5)^2}=6\) \(\text { Area of rectangle }(A)=A B \times B C\) \(=14 \times 6\) \(\mathrm{~A}=84\)
JEE Main 09.04.2019
Conic Section
119790
The sum of the squares of the lengths of the chords intercepted on the circle, \(x^2+y^2=16\), by the lines, \(x+y=n, n \in N\), where \(N\) is the set of all natural numbers, is
1 320
2 105
3 160
4 210
Explanation:
D Equation of given circle, \(x^2+y^2=16\) Equation of chord \(\mathrm{x}+\mathrm{y}=\mathrm{n} \quad\) when, \(\mathrm{n} \in \mathrm{N}\) Length of chord \(=2 \sqrt{16-\frac{n^2}{2}}\) Squaring on both sides, (length of chord \()^2=2\left(32-n^2\right)\) Possible value of \(n=1,2,3,4,5\) Sum of the squares of the length of the chord \(=2(31)+2(28)+2(23)+2(16)+2(7)\) \(=210\)
119786
If one of the diameters of the circle, given by the equation, \(x^2+y^2-4 x+6 y-12=0\), is a chord of a circle \(S\), whose centre is at \((-3,2)\) then the radius of \(S\) is
1 \(5 \sqrt{2}\)
2 \(5 \sqrt{3}\)
3 5
4 10
Explanation:
B Given circle equation- \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}-12=0\) \(\text { From comparing general equation of circle we get }\) \(\quad \mathrm{g}_1=+2, \mathrm{f},=-3 \quad \mathrm{c}_1=-12\) \(\mathrm{r}_1=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}_1}\) \(\mathrm{r}_1=\sqrt{4+9+12}=5\) \(\mathrm{C}_2=(-3,2)\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(5)^2+(5)^2}\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{50}\) \(\mathrm{C}_2 \mathrm{~A}=\sqrt{(5)^2+(\sqrt{50})^2}=\sqrt{75}\) \(\mathrm{C}_2 \mathrm{~A}=\mathrm{r}=5 \sqrt{3}\)
JEE Main-2016
Conic Section
119787
A circle cuts a chord of length 4 a on the \(\mathrm{X}\)-axis and passes through a point on the \(\mathrm{Y}\)-axis, distant \(2 \mathrm{~b}\) from the origin. Then the locus of the centre of this circle, is
1 a parabola
2 an ellipse
3 a straight line
4 a hyperbola
Explanation:
A Let equation of the circle is, \(x^2+y^2+2 g x+2 f y+c=0\) Center \(=(-\mathrm{g},-\mathrm{f})\) Radius \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) The circle equation passes through \((0,2 b)\) \(4 \mathrm{~b}^2+4 \mathrm{fb}+\mathrm{c}=0\) intercept on \(\mathrm{x}-\) axis \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}=4 \mathrm{a}\) \(\mathrm{g}^2-\mathrm{c}=4 \mathrm{a}^2\) \(\mathrm{c}=\mathrm{g}^2-4 \mathrm{a}^2\) Putting in eq - (i) \(4 b^2+4 f+g^2-4 a^2=0\) \(x^2+4 y+4\left(b^2-a^2\right)=0\)Its represent parabola.
JEE Main 11.01.2019
Conic Section
119788
A square is inscribed in the circle \(x^2+y^2-6 x+\) \(8 y-103=0\) with its sides parallel to the coordinate axes. Then, the distance of the vertex of this square which is nearest to the origin is
1 6
2 13
3 \(\sqrt{41}\)
4 \(\sqrt{137}\)
Explanation:
C Given, \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+8 \mathrm{y}-103=0\) \((-\mathrm{g},-\mathrm{f})=3,-4\) \(\mathrm{c}=-103\) \(\text { From figure, }\) \(\mathrm{R}=\sqrt{9+16+103}=8 \sqrt{2}\) \(\mathrm{OB}=\sqrt{265}, \mathrm{OC}=\sqrt{137}, \mathrm{OD}=\sqrt{41}\) \(\text { Hence OD }=\sqrt{41} \text { is the nearest to the origin }\) From figure, Hence OD \(=\sqrt{41}\) is the nearest to the origin
JEE Main 11.01.2019
Conic Section
119789
A rectangle is inscribed in a circle with a diameter lying along the line \(3 y=x+7\). If the two adjacent vertices of the rectangle are \((-8\), 5 ) and \((6,5)\), then the area of the rectangle (in sq units) is
1 72
2 84
3 98
4 56
Explanation:
B From figure, \(\text { Let vertex } \mathrm{C} \text { is }(6, \mathrm{k})\) \(\therefore \text { centre of circle is }\) \(\left(\frac{-8+6}{2}, \frac{5+\mathrm{k}}{2}\right)=\left(-1, \frac{5+\mathrm{k}}{2}\right)\) Which is lying on diameter of circle \(3 \mathrm{y}=\mathrm{x}+7\) \(3\left(\frac{5+\mathrm{k}}{2}\right)=-1+7\) \(15+3 \mathrm{k}=12\) \(\mathrm{k}=-1\) \(\text { vertex of } \mathrm{C}=(6,-1)\) \(\text { Length of } A B=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(=\sqrt{(6+8)^2+(5-5)^2}=14\) \(\text { Length of } B C=\sqrt{(6-6)^2+(-1-5)^2}=6\) \(\text { Area of rectangle }(A)=A B \times B C\) \(=14 \times 6\) \(\mathrm{~A}=84\)
JEE Main 09.04.2019
Conic Section
119790
The sum of the squares of the lengths of the chords intercepted on the circle, \(x^2+y^2=16\), by the lines, \(x+y=n, n \in N\), where \(N\) is the set of all natural numbers, is
1 320
2 105
3 160
4 210
Explanation:
D Equation of given circle, \(x^2+y^2=16\) Equation of chord \(\mathrm{x}+\mathrm{y}=\mathrm{n} \quad\) when, \(\mathrm{n} \in \mathrm{N}\) Length of chord \(=2 \sqrt{16-\frac{n^2}{2}}\) Squaring on both sides, (length of chord \()^2=2\left(32-n^2\right)\) Possible value of \(n=1,2,3,4,5\) Sum of the squares of the length of the chord \(=2(31)+2(28)+2(23)+2(16)+2(7)\) \(=210\)
119786
If one of the diameters of the circle, given by the equation, \(x^2+y^2-4 x+6 y-12=0\), is a chord of a circle \(S\), whose centre is at \((-3,2)\) then the radius of \(S\) is
1 \(5 \sqrt{2}\)
2 \(5 \sqrt{3}\)
3 5
4 10
Explanation:
B Given circle equation- \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}-12=0\) \(\text { From comparing general equation of circle we get }\) \(\quad \mathrm{g}_1=+2, \mathrm{f},=-3 \quad \mathrm{c}_1=-12\) \(\mathrm{r}_1=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}_1}\) \(\mathrm{r}_1=\sqrt{4+9+12}=5\) \(\mathrm{C}_2=(-3,2)\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(5)^2+(5)^2}\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{50}\) \(\mathrm{C}_2 \mathrm{~A}=\sqrt{(5)^2+(\sqrt{50})^2}=\sqrt{75}\) \(\mathrm{C}_2 \mathrm{~A}=\mathrm{r}=5 \sqrt{3}\)
JEE Main-2016
Conic Section
119787
A circle cuts a chord of length 4 a on the \(\mathrm{X}\)-axis and passes through a point on the \(\mathrm{Y}\)-axis, distant \(2 \mathrm{~b}\) from the origin. Then the locus of the centre of this circle, is
1 a parabola
2 an ellipse
3 a straight line
4 a hyperbola
Explanation:
A Let equation of the circle is, \(x^2+y^2+2 g x+2 f y+c=0\) Center \(=(-\mathrm{g},-\mathrm{f})\) Radius \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) The circle equation passes through \((0,2 b)\) \(4 \mathrm{~b}^2+4 \mathrm{fb}+\mathrm{c}=0\) intercept on \(\mathrm{x}-\) axis \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}=4 \mathrm{a}\) \(\mathrm{g}^2-\mathrm{c}=4 \mathrm{a}^2\) \(\mathrm{c}=\mathrm{g}^2-4 \mathrm{a}^2\) Putting in eq - (i) \(4 b^2+4 f+g^2-4 a^2=0\) \(x^2+4 y+4\left(b^2-a^2\right)=0\)Its represent parabola.
JEE Main 11.01.2019
Conic Section
119788
A square is inscribed in the circle \(x^2+y^2-6 x+\) \(8 y-103=0\) with its sides parallel to the coordinate axes. Then, the distance of the vertex of this square which is nearest to the origin is
1 6
2 13
3 \(\sqrt{41}\)
4 \(\sqrt{137}\)
Explanation:
C Given, \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+8 \mathrm{y}-103=0\) \((-\mathrm{g},-\mathrm{f})=3,-4\) \(\mathrm{c}=-103\) \(\text { From figure, }\) \(\mathrm{R}=\sqrt{9+16+103}=8 \sqrt{2}\) \(\mathrm{OB}=\sqrt{265}, \mathrm{OC}=\sqrt{137}, \mathrm{OD}=\sqrt{41}\) \(\text { Hence OD }=\sqrt{41} \text { is the nearest to the origin }\) From figure, Hence OD \(=\sqrt{41}\) is the nearest to the origin
JEE Main 11.01.2019
Conic Section
119789
A rectangle is inscribed in a circle with a diameter lying along the line \(3 y=x+7\). If the two adjacent vertices of the rectangle are \((-8\), 5 ) and \((6,5)\), then the area of the rectangle (in sq units) is
1 72
2 84
3 98
4 56
Explanation:
B From figure, \(\text { Let vertex } \mathrm{C} \text { is }(6, \mathrm{k})\) \(\therefore \text { centre of circle is }\) \(\left(\frac{-8+6}{2}, \frac{5+\mathrm{k}}{2}\right)=\left(-1, \frac{5+\mathrm{k}}{2}\right)\) Which is lying on diameter of circle \(3 \mathrm{y}=\mathrm{x}+7\) \(3\left(\frac{5+\mathrm{k}}{2}\right)=-1+7\) \(15+3 \mathrm{k}=12\) \(\mathrm{k}=-1\) \(\text { vertex of } \mathrm{C}=(6,-1)\) \(\text { Length of } A B=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(=\sqrt{(6+8)^2+(5-5)^2}=14\) \(\text { Length of } B C=\sqrt{(6-6)^2+(-1-5)^2}=6\) \(\text { Area of rectangle }(A)=A B \times B C\) \(=14 \times 6\) \(\mathrm{~A}=84\)
JEE Main 09.04.2019
Conic Section
119790
The sum of the squares of the lengths of the chords intercepted on the circle, \(x^2+y^2=16\), by the lines, \(x+y=n, n \in N\), where \(N\) is the set of all natural numbers, is
1 320
2 105
3 160
4 210
Explanation:
D Equation of given circle, \(x^2+y^2=16\) Equation of chord \(\mathrm{x}+\mathrm{y}=\mathrm{n} \quad\) when, \(\mathrm{n} \in \mathrm{N}\) Length of chord \(=2 \sqrt{16-\frac{n^2}{2}}\) Squaring on both sides, (length of chord \()^2=2\left(32-n^2\right)\) Possible value of \(n=1,2,3,4,5\) Sum of the squares of the length of the chord \(=2(31)+2(28)+2(23)+2(16)+2(7)\) \(=210\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119786
If one of the diameters of the circle, given by the equation, \(x^2+y^2-4 x+6 y-12=0\), is a chord of a circle \(S\), whose centre is at \((-3,2)\) then the radius of \(S\) is
1 \(5 \sqrt{2}\)
2 \(5 \sqrt{3}\)
3 5
4 10
Explanation:
B Given circle equation- \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}-12=0\) \(\text { From comparing general equation of circle we get }\) \(\quad \mathrm{g}_1=+2, \mathrm{f},=-3 \quad \mathrm{c}_1=-12\) \(\mathrm{r}_1=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}_1}\) \(\mathrm{r}_1=\sqrt{4+9+12}=5\) \(\mathrm{C}_2=(-3,2)\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(5)^2+(5)^2}\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{50}\) \(\mathrm{C}_2 \mathrm{~A}=\sqrt{(5)^2+(\sqrt{50})^2}=\sqrt{75}\) \(\mathrm{C}_2 \mathrm{~A}=\mathrm{r}=5 \sqrt{3}\)
JEE Main-2016
Conic Section
119787
A circle cuts a chord of length 4 a on the \(\mathrm{X}\)-axis and passes through a point on the \(\mathrm{Y}\)-axis, distant \(2 \mathrm{~b}\) from the origin. Then the locus of the centre of this circle, is
1 a parabola
2 an ellipse
3 a straight line
4 a hyperbola
Explanation:
A Let equation of the circle is, \(x^2+y^2+2 g x+2 f y+c=0\) Center \(=(-\mathrm{g},-\mathrm{f})\) Radius \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) The circle equation passes through \((0,2 b)\) \(4 \mathrm{~b}^2+4 \mathrm{fb}+\mathrm{c}=0\) intercept on \(\mathrm{x}-\) axis \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}=4 \mathrm{a}\) \(\mathrm{g}^2-\mathrm{c}=4 \mathrm{a}^2\) \(\mathrm{c}=\mathrm{g}^2-4 \mathrm{a}^2\) Putting in eq - (i) \(4 b^2+4 f+g^2-4 a^2=0\) \(x^2+4 y+4\left(b^2-a^2\right)=0\)Its represent parabola.
JEE Main 11.01.2019
Conic Section
119788
A square is inscribed in the circle \(x^2+y^2-6 x+\) \(8 y-103=0\) with its sides parallel to the coordinate axes. Then, the distance of the vertex of this square which is nearest to the origin is
1 6
2 13
3 \(\sqrt{41}\)
4 \(\sqrt{137}\)
Explanation:
C Given, \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+8 \mathrm{y}-103=0\) \((-\mathrm{g},-\mathrm{f})=3,-4\) \(\mathrm{c}=-103\) \(\text { From figure, }\) \(\mathrm{R}=\sqrt{9+16+103}=8 \sqrt{2}\) \(\mathrm{OB}=\sqrt{265}, \mathrm{OC}=\sqrt{137}, \mathrm{OD}=\sqrt{41}\) \(\text { Hence OD }=\sqrt{41} \text { is the nearest to the origin }\) From figure, Hence OD \(=\sqrt{41}\) is the nearest to the origin
JEE Main 11.01.2019
Conic Section
119789
A rectangle is inscribed in a circle with a diameter lying along the line \(3 y=x+7\). If the two adjacent vertices of the rectangle are \((-8\), 5 ) and \((6,5)\), then the area of the rectangle (in sq units) is
1 72
2 84
3 98
4 56
Explanation:
B From figure, \(\text { Let vertex } \mathrm{C} \text { is }(6, \mathrm{k})\) \(\therefore \text { centre of circle is }\) \(\left(\frac{-8+6}{2}, \frac{5+\mathrm{k}}{2}\right)=\left(-1, \frac{5+\mathrm{k}}{2}\right)\) Which is lying on diameter of circle \(3 \mathrm{y}=\mathrm{x}+7\) \(3\left(\frac{5+\mathrm{k}}{2}\right)=-1+7\) \(15+3 \mathrm{k}=12\) \(\mathrm{k}=-1\) \(\text { vertex of } \mathrm{C}=(6,-1)\) \(\text { Length of } A B=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(=\sqrt{(6+8)^2+(5-5)^2}=14\) \(\text { Length of } B C=\sqrt{(6-6)^2+(-1-5)^2}=6\) \(\text { Area of rectangle }(A)=A B \times B C\) \(=14 \times 6\) \(\mathrm{~A}=84\)
JEE Main 09.04.2019
Conic Section
119790
The sum of the squares of the lengths of the chords intercepted on the circle, \(x^2+y^2=16\), by the lines, \(x+y=n, n \in N\), where \(N\) is the set of all natural numbers, is
1 320
2 105
3 160
4 210
Explanation:
D Equation of given circle, \(x^2+y^2=16\) Equation of chord \(\mathrm{x}+\mathrm{y}=\mathrm{n} \quad\) when, \(\mathrm{n} \in \mathrm{N}\) Length of chord \(=2 \sqrt{16-\frac{n^2}{2}}\) Squaring on both sides, (length of chord \()^2=2\left(32-n^2\right)\) Possible value of \(n=1,2,3,4,5\) Sum of the squares of the length of the chord \(=2(31)+2(28)+2(23)+2(16)+2(7)\) \(=210\)
119786
If one of the diameters of the circle, given by the equation, \(x^2+y^2-4 x+6 y-12=0\), is a chord of a circle \(S\), whose centre is at \((-3,2)\) then the radius of \(S\) is
1 \(5 \sqrt{2}\)
2 \(5 \sqrt{3}\)
3 5
4 10
Explanation:
B Given circle equation- \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}-12=0\) \(\text { From comparing general equation of circle we get }\) \(\quad \mathrm{g}_1=+2, \mathrm{f},=-3 \quad \mathrm{c}_1=-12\) \(\mathrm{r}_1=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}_1}\) \(\mathrm{r}_1=\sqrt{4+9+12}=5\) \(\mathrm{C}_2=(-3,2)\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(5)^2+(5)^2}\) \(\mathrm{C}_1 \mathrm{C}_2=\sqrt{50}\) \(\mathrm{C}_2 \mathrm{~A}=\sqrt{(5)^2+(\sqrt{50})^2}=\sqrt{75}\) \(\mathrm{C}_2 \mathrm{~A}=\mathrm{r}=5 \sqrt{3}\)
JEE Main-2016
Conic Section
119787
A circle cuts a chord of length 4 a on the \(\mathrm{X}\)-axis and passes through a point on the \(\mathrm{Y}\)-axis, distant \(2 \mathrm{~b}\) from the origin. Then the locus of the centre of this circle, is
1 a parabola
2 an ellipse
3 a straight line
4 a hyperbola
Explanation:
A Let equation of the circle is, \(x^2+y^2+2 g x+2 f y+c=0\) Center \(=(-\mathrm{g},-\mathrm{f})\) Radius \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) The circle equation passes through \((0,2 b)\) \(4 \mathrm{~b}^2+4 \mathrm{fb}+\mathrm{c}=0\) intercept on \(\mathrm{x}-\) axis \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}=4 \mathrm{a}\) \(\mathrm{g}^2-\mathrm{c}=4 \mathrm{a}^2\) \(\mathrm{c}=\mathrm{g}^2-4 \mathrm{a}^2\) Putting in eq - (i) \(4 b^2+4 f+g^2-4 a^2=0\) \(x^2+4 y+4\left(b^2-a^2\right)=0\)Its represent parabola.
JEE Main 11.01.2019
Conic Section
119788
A square is inscribed in the circle \(x^2+y^2-6 x+\) \(8 y-103=0\) with its sides parallel to the coordinate axes. Then, the distance of the vertex of this square which is nearest to the origin is
1 6
2 13
3 \(\sqrt{41}\)
4 \(\sqrt{137}\)
Explanation:
C Given, \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+8 \mathrm{y}-103=0\) \((-\mathrm{g},-\mathrm{f})=3,-4\) \(\mathrm{c}=-103\) \(\text { From figure, }\) \(\mathrm{R}=\sqrt{9+16+103}=8 \sqrt{2}\) \(\mathrm{OB}=\sqrt{265}, \mathrm{OC}=\sqrt{137}, \mathrm{OD}=\sqrt{41}\) \(\text { Hence OD }=\sqrt{41} \text { is the nearest to the origin }\) From figure, Hence OD \(=\sqrt{41}\) is the nearest to the origin
JEE Main 11.01.2019
Conic Section
119789
A rectangle is inscribed in a circle with a diameter lying along the line \(3 y=x+7\). If the two adjacent vertices of the rectangle are \((-8\), 5 ) and \((6,5)\), then the area of the rectangle (in sq units) is
1 72
2 84
3 98
4 56
Explanation:
B From figure, \(\text { Let vertex } \mathrm{C} \text { is }(6, \mathrm{k})\) \(\therefore \text { centre of circle is }\) \(\left(\frac{-8+6}{2}, \frac{5+\mathrm{k}}{2}\right)=\left(-1, \frac{5+\mathrm{k}}{2}\right)\) Which is lying on diameter of circle \(3 \mathrm{y}=\mathrm{x}+7\) \(3\left(\frac{5+\mathrm{k}}{2}\right)=-1+7\) \(15+3 \mathrm{k}=12\) \(\mathrm{k}=-1\) \(\text { vertex of } \mathrm{C}=(6,-1)\) \(\text { Length of } A B=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(=\sqrt{(6+8)^2+(5-5)^2}=14\) \(\text { Length of } B C=\sqrt{(6-6)^2+(-1-5)^2}=6\) \(\text { Area of rectangle }(A)=A B \times B C\) \(=14 \times 6\) \(\mathrm{~A}=84\)
JEE Main 09.04.2019
Conic Section
119790
The sum of the squares of the lengths of the chords intercepted on the circle, \(x^2+y^2=16\), by the lines, \(x+y=n, n \in N\), where \(N\) is the set of all natural numbers, is
1 320
2 105
3 160
4 210
Explanation:
D Equation of given circle, \(x^2+y^2=16\) Equation of chord \(\mathrm{x}+\mathrm{y}=\mathrm{n} \quad\) when, \(\mathrm{n} \in \mathrm{N}\) Length of chord \(=2 \sqrt{16-\frac{n^2}{2}}\) Squaring on both sides, (length of chord \()^2=2\left(32-n^2\right)\) Possible value of \(n=1,2,3,4,5\) Sum of the squares of the length of the chord \(=2(31)+2(28)+2(23)+2(16)+2(7)\) \(=210\)
