119778
The radius of a circle whose center is \((2,1)\) and one of the chords is a diameter of the circle \(x^2+\) \(y^2-2 x-6 y+6=0\) is \(\qquad\) units.
1 3
2 4
3 2
4 1
Explanation:
A : Centre of the given circle is \(C(1,3)\) and \(r=\sqrt{1+9-6}=2\) Let, \(\mathrm{D}(2,1)\) be the centre of the circle clearly \(\mathrm{C}\) is the mid-point of chord \(\mathrm{AB}\) \(\therefore \mathrm{CD} \perp \mathrm{AB}\). \(\therefore\) From the right angled \(\triangle \mathrm{ACD}, \mathrm{AD}^2=\mathrm{AC}^2+\mathrm{CD}^2\) i.e. \(\mathrm{AD}^2=2^2+(2-1)^2+(1-3)^2=4+1+4=9\) \(\mathrm{AD}=\sqrt{9}=3\) \(\mathrm{AD}=3\) is the required radius.
AP EAMCET-23.08.2021
Conic Section
119779
The straight line \(x \cos \alpha+y \sin \alpha=p\) cuts the circle \(x^2+y^2-a^2=0\) at \(A\) and \(B\). Then the equation of circle having \(A B\) as diameter is
D Given straight line equation, \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\) and circle equation- \(x^2+y^2=a^2\) \(x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0\) \(\text { Centre of circle }=\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p}=0-(\) diameter \()\) Since, it is passes through the point \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) Then, \(\cos \alpha\left(-\frac{\lambda \cos \alpha}{2}\right)+\left(\sin \alpha \frac{-\lambda \sin \alpha}{2}\right)-p=0\) \(\frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}-p=0\) \(p=-\frac{\lambda}{2}\left(\cos ^2 \alpha+\sin ^2 \alpha\right)\) \(p=\frac{-\lambda}{2}\) \(\lambda=-2 p\) Then, the equation of circle is- \(x^2+y^2-a^2-2 p(x \cos \alpha+y \sin \alpha-p)=0\)
AP EAMCET-08.07.2022
Conic Section
119780
If the area of an equilateral triangle inscribed in the circle, \(x^2+y^2+10 x+12 y+c=0\) is \(27 \sqrt{3}\) sq units, then \(c\) is equal to
1 20
2 -25
3 13
4 25
Explanation:
D Given circle equation\(\mathrm{x}^2+\mathrm{y}^2+10 \mathrm{x}+12 \mathrm{y}+\mathrm{c}=0\) is \(27 \sqrt{3}\) square units. Area of the equilateral triangle inscribed in the circle \(\therefore \quad \frac{\sqrt{3}}{2} a^2=27 \sqrt{3} \Rightarrow \mathrm{a}=6 \sqrt{3}\) From question, \(\angle \mathrm{OAD}=30^{\circ}\) Then, \(\quad r \cos 30^{\circ}=\frac{6 \sqrt{3}}{2}\) \(\mathrm{r}=6\) \(6=\sqrt{25+36-\mathrm{c}}\) So, \(\quad \mathrm{c}=25\)
JEE Main 10.01.2019
Conic Section
119781
If a circle passes through the point \((a, b)\) and cuts the circle \(x^2+y^2=p^2\) orthogonally, then the equation of the locus of its centre is
1 \(2 a x+2 b y-\left(a^2+b^2+p^2\right)=0\)
2 \(x^2+y^2-2 a x-3 b y+\left(a^2-b^2-p^2\right)=0\)
3 \(2 a x+2 b y-\left(a^2-b^2+p^2\right)=0\)
4 \(x^2+y^2-3 a x-4 b y+\left(a^2+b^2-p^2\right)=0\)
Explanation:
A Let the equation of the circle \((a, b)\) be \(x^2+y^2+2 g x+2 f y+c=0\) This circle passes through the point \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ga}+2 \mathrm{fb}+\mathrm{c}=0\) If circle (1) cuts the circle \(x^2+y^2-p^2=0\) orthogonally, then \(2 \times \mathrm{g} \times 0+2 \mathrm{f} \times 0=\mathrm{c}-\mathrm{p}^2\) \(\mathrm{c}=\mathrm{p}^2\) If \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is the centre of the circle (1), then \(\mathrm{h}=-\mathrm{g} \Rightarrow \mathrm{g}=-\mathrm{h}\) \(\mathrm{k}=-\mathrm{f} \Rightarrow \mathrm{f}=-\mathrm{k}\) substituting in (2), we get - Or \(a^2+b^2-2 a h-2 b k+p^2=0\) \(2 a h+2 b k-\left(a^2+b^2+p^2\right)=0\) Therefore, Locus of \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is \(2 a x+2 b y-\left(a^2+b^2+p^2\right)=0\)
119778
The radius of a circle whose center is \((2,1)\) and one of the chords is a diameter of the circle \(x^2+\) \(y^2-2 x-6 y+6=0\) is \(\qquad\) units.
1 3
2 4
3 2
4 1
Explanation:
A : Centre of the given circle is \(C(1,3)\) and \(r=\sqrt{1+9-6}=2\) Let, \(\mathrm{D}(2,1)\) be the centre of the circle clearly \(\mathrm{C}\) is the mid-point of chord \(\mathrm{AB}\) \(\therefore \mathrm{CD} \perp \mathrm{AB}\). \(\therefore\) From the right angled \(\triangle \mathrm{ACD}, \mathrm{AD}^2=\mathrm{AC}^2+\mathrm{CD}^2\) i.e. \(\mathrm{AD}^2=2^2+(2-1)^2+(1-3)^2=4+1+4=9\) \(\mathrm{AD}=\sqrt{9}=3\) \(\mathrm{AD}=3\) is the required radius.
AP EAMCET-23.08.2021
Conic Section
119779
The straight line \(x \cos \alpha+y \sin \alpha=p\) cuts the circle \(x^2+y^2-a^2=0\) at \(A\) and \(B\). Then the equation of circle having \(A B\) as diameter is
D Given straight line equation, \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\) and circle equation- \(x^2+y^2=a^2\) \(x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0\) \(\text { Centre of circle }=\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p}=0-(\) diameter \()\) Since, it is passes through the point \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) Then, \(\cos \alpha\left(-\frac{\lambda \cos \alpha}{2}\right)+\left(\sin \alpha \frac{-\lambda \sin \alpha}{2}\right)-p=0\) \(\frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}-p=0\) \(p=-\frac{\lambda}{2}\left(\cos ^2 \alpha+\sin ^2 \alpha\right)\) \(p=\frac{-\lambda}{2}\) \(\lambda=-2 p\) Then, the equation of circle is- \(x^2+y^2-a^2-2 p(x \cos \alpha+y \sin \alpha-p)=0\)
AP EAMCET-08.07.2022
Conic Section
119780
If the area of an equilateral triangle inscribed in the circle, \(x^2+y^2+10 x+12 y+c=0\) is \(27 \sqrt{3}\) sq units, then \(c\) is equal to
1 20
2 -25
3 13
4 25
Explanation:
D Given circle equation\(\mathrm{x}^2+\mathrm{y}^2+10 \mathrm{x}+12 \mathrm{y}+\mathrm{c}=0\) is \(27 \sqrt{3}\) square units. Area of the equilateral triangle inscribed in the circle \(\therefore \quad \frac{\sqrt{3}}{2} a^2=27 \sqrt{3} \Rightarrow \mathrm{a}=6 \sqrt{3}\) From question, \(\angle \mathrm{OAD}=30^{\circ}\) Then, \(\quad r \cos 30^{\circ}=\frac{6 \sqrt{3}}{2}\) \(\mathrm{r}=6\) \(6=\sqrt{25+36-\mathrm{c}}\) So, \(\quad \mathrm{c}=25\)
JEE Main 10.01.2019
Conic Section
119781
If a circle passes through the point \((a, b)\) and cuts the circle \(x^2+y^2=p^2\) orthogonally, then the equation of the locus of its centre is
1 \(2 a x+2 b y-\left(a^2+b^2+p^2\right)=0\)
2 \(x^2+y^2-2 a x-3 b y+\left(a^2-b^2-p^2\right)=0\)
3 \(2 a x+2 b y-\left(a^2-b^2+p^2\right)=0\)
4 \(x^2+y^2-3 a x-4 b y+\left(a^2+b^2-p^2\right)=0\)
Explanation:
A Let the equation of the circle \((a, b)\) be \(x^2+y^2+2 g x+2 f y+c=0\) This circle passes through the point \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ga}+2 \mathrm{fb}+\mathrm{c}=0\) If circle (1) cuts the circle \(x^2+y^2-p^2=0\) orthogonally, then \(2 \times \mathrm{g} \times 0+2 \mathrm{f} \times 0=\mathrm{c}-\mathrm{p}^2\) \(\mathrm{c}=\mathrm{p}^2\) If \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is the centre of the circle (1), then \(\mathrm{h}=-\mathrm{g} \Rightarrow \mathrm{g}=-\mathrm{h}\) \(\mathrm{k}=-\mathrm{f} \Rightarrow \mathrm{f}=-\mathrm{k}\) substituting in (2), we get - Or \(a^2+b^2-2 a h-2 b k+p^2=0\) \(2 a h+2 b k-\left(a^2+b^2+p^2\right)=0\) Therefore, Locus of \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is \(2 a x+2 b y-\left(a^2+b^2+p^2\right)=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119778
The radius of a circle whose center is \((2,1)\) and one of the chords is a diameter of the circle \(x^2+\) \(y^2-2 x-6 y+6=0\) is \(\qquad\) units.
1 3
2 4
3 2
4 1
Explanation:
A : Centre of the given circle is \(C(1,3)\) and \(r=\sqrt{1+9-6}=2\) Let, \(\mathrm{D}(2,1)\) be the centre of the circle clearly \(\mathrm{C}\) is the mid-point of chord \(\mathrm{AB}\) \(\therefore \mathrm{CD} \perp \mathrm{AB}\). \(\therefore\) From the right angled \(\triangle \mathrm{ACD}, \mathrm{AD}^2=\mathrm{AC}^2+\mathrm{CD}^2\) i.e. \(\mathrm{AD}^2=2^2+(2-1)^2+(1-3)^2=4+1+4=9\) \(\mathrm{AD}=\sqrt{9}=3\) \(\mathrm{AD}=3\) is the required radius.
AP EAMCET-23.08.2021
Conic Section
119779
The straight line \(x \cos \alpha+y \sin \alpha=p\) cuts the circle \(x^2+y^2-a^2=0\) at \(A\) and \(B\). Then the equation of circle having \(A B\) as diameter is
D Given straight line equation, \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\) and circle equation- \(x^2+y^2=a^2\) \(x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0\) \(\text { Centre of circle }=\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p}=0-(\) diameter \()\) Since, it is passes through the point \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) Then, \(\cos \alpha\left(-\frac{\lambda \cos \alpha}{2}\right)+\left(\sin \alpha \frac{-\lambda \sin \alpha}{2}\right)-p=0\) \(\frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}-p=0\) \(p=-\frac{\lambda}{2}\left(\cos ^2 \alpha+\sin ^2 \alpha\right)\) \(p=\frac{-\lambda}{2}\) \(\lambda=-2 p\) Then, the equation of circle is- \(x^2+y^2-a^2-2 p(x \cos \alpha+y \sin \alpha-p)=0\)
AP EAMCET-08.07.2022
Conic Section
119780
If the area of an equilateral triangle inscribed in the circle, \(x^2+y^2+10 x+12 y+c=0\) is \(27 \sqrt{3}\) sq units, then \(c\) is equal to
1 20
2 -25
3 13
4 25
Explanation:
D Given circle equation\(\mathrm{x}^2+\mathrm{y}^2+10 \mathrm{x}+12 \mathrm{y}+\mathrm{c}=0\) is \(27 \sqrt{3}\) square units. Area of the equilateral triangle inscribed in the circle \(\therefore \quad \frac{\sqrt{3}}{2} a^2=27 \sqrt{3} \Rightarrow \mathrm{a}=6 \sqrt{3}\) From question, \(\angle \mathrm{OAD}=30^{\circ}\) Then, \(\quad r \cos 30^{\circ}=\frac{6 \sqrt{3}}{2}\) \(\mathrm{r}=6\) \(6=\sqrt{25+36-\mathrm{c}}\) So, \(\quad \mathrm{c}=25\)
JEE Main 10.01.2019
Conic Section
119781
If a circle passes through the point \((a, b)\) and cuts the circle \(x^2+y^2=p^2\) orthogonally, then the equation of the locus of its centre is
1 \(2 a x+2 b y-\left(a^2+b^2+p^2\right)=0\)
2 \(x^2+y^2-2 a x-3 b y+\left(a^2-b^2-p^2\right)=0\)
3 \(2 a x+2 b y-\left(a^2-b^2+p^2\right)=0\)
4 \(x^2+y^2-3 a x-4 b y+\left(a^2+b^2-p^2\right)=0\)
Explanation:
A Let the equation of the circle \((a, b)\) be \(x^2+y^2+2 g x+2 f y+c=0\) This circle passes through the point \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ga}+2 \mathrm{fb}+\mathrm{c}=0\) If circle (1) cuts the circle \(x^2+y^2-p^2=0\) orthogonally, then \(2 \times \mathrm{g} \times 0+2 \mathrm{f} \times 0=\mathrm{c}-\mathrm{p}^2\) \(\mathrm{c}=\mathrm{p}^2\) If \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is the centre of the circle (1), then \(\mathrm{h}=-\mathrm{g} \Rightarrow \mathrm{g}=-\mathrm{h}\) \(\mathrm{k}=-\mathrm{f} \Rightarrow \mathrm{f}=-\mathrm{k}\) substituting in (2), we get - Or \(a^2+b^2-2 a h-2 b k+p^2=0\) \(2 a h+2 b k-\left(a^2+b^2+p^2\right)=0\) Therefore, Locus of \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is \(2 a x+2 b y-\left(a^2+b^2+p^2\right)=0\)
119778
The radius of a circle whose center is \((2,1)\) and one of the chords is a diameter of the circle \(x^2+\) \(y^2-2 x-6 y+6=0\) is \(\qquad\) units.
1 3
2 4
3 2
4 1
Explanation:
A : Centre of the given circle is \(C(1,3)\) and \(r=\sqrt{1+9-6}=2\) Let, \(\mathrm{D}(2,1)\) be the centre of the circle clearly \(\mathrm{C}\) is the mid-point of chord \(\mathrm{AB}\) \(\therefore \mathrm{CD} \perp \mathrm{AB}\). \(\therefore\) From the right angled \(\triangle \mathrm{ACD}, \mathrm{AD}^2=\mathrm{AC}^2+\mathrm{CD}^2\) i.e. \(\mathrm{AD}^2=2^2+(2-1)^2+(1-3)^2=4+1+4=9\) \(\mathrm{AD}=\sqrt{9}=3\) \(\mathrm{AD}=3\) is the required radius.
AP EAMCET-23.08.2021
Conic Section
119779
The straight line \(x \cos \alpha+y \sin \alpha=p\) cuts the circle \(x^2+y^2-a^2=0\) at \(A\) and \(B\). Then the equation of circle having \(A B\) as diameter is
D Given straight line equation, \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\) and circle equation- \(x^2+y^2=a^2\) \(x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0\) \(\text { Centre of circle }=\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p}=0-(\) diameter \()\) Since, it is passes through the point \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) Then, \(\cos \alpha\left(-\frac{\lambda \cos \alpha}{2}\right)+\left(\sin \alpha \frac{-\lambda \sin \alpha}{2}\right)-p=0\) \(\frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}-p=0\) \(p=-\frac{\lambda}{2}\left(\cos ^2 \alpha+\sin ^2 \alpha\right)\) \(p=\frac{-\lambda}{2}\) \(\lambda=-2 p\) Then, the equation of circle is- \(x^2+y^2-a^2-2 p(x \cos \alpha+y \sin \alpha-p)=0\)
AP EAMCET-08.07.2022
Conic Section
119780
If the area of an equilateral triangle inscribed in the circle, \(x^2+y^2+10 x+12 y+c=0\) is \(27 \sqrt{3}\) sq units, then \(c\) is equal to
1 20
2 -25
3 13
4 25
Explanation:
D Given circle equation\(\mathrm{x}^2+\mathrm{y}^2+10 \mathrm{x}+12 \mathrm{y}+\mathrm{c}=0\) is \(27 \sqrt{3}\) square units. Area of the equilateral triangle inscribed in the circle \(\therefore \quad \frac{\sqrt{3}}{2} a^2=27 \sqrt{3} \Rightarrow \mathrm{a}=6 \sqrt{3}\) From question, \(\angle \mathrm{OAD}=30^{\circ}\) Then, \(\quad r \cos 30^{\circ}=\frac{6 \sqrt{3}}{2}\) \(\mathrm{r}=6\) \(6=\sqrt{25+36-\mathrm{c}}\) So, \(\quad \mathrm{c}=25\)
JEE Main 10.01.2019
Conic Section
119781
If a circle passes through the point \((a, b)\) and cuts the circle \(x^2+y^2=p^2\) orthogonally, then the equation of the locus of its centre is
1 \(2 a x+2 b y-\left(a^2+b^2+p^2\right)=0\)
2 \(x^2+y^2-2 a x-3 b y+\left(a^2-b^2-p^2\right)=0\)
3 \(2 a x+2 b y-\left(a^2-b^2+p^2\right)=0\)
4 \(x^2+y^2-3 a x-4 b y+\left(a^2+b^2-p^2\right)=0\)
Explanation:
A Let the equation of the circle \((a, b)\) be \(x^2+y^2+2 g x+2 f y+c=0\) This circle passes through the point \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ga}+2 \mathrm{fb}+\mathrm{c}=0\) If circle (1) cuts the circle \(x^2+y^2-p^2=0\) orthogonally, then \(2 \times \mathrm{g} \times 0+2 \mathrm{f} \times 0=\mathrm{c}-\mathrm{p}^2\) \(\mathrm{c}=\mathrm{p}^2\) If \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is the centre of the circle (1), then \(\mathrm{h}=-\mathrm{g} \Rightarrow \mathrm{g}=-\mathrm{h}\) \(\mathrm{k}=-\mathrm{f} \Rightarrow \mathrm{f}=-\mathrm{k}\) substituting in (2), we get - Or \(a^2+b^2-2 a h-2 b k+p^2=0\) \(2 a h+2 b k-\left(a^2+b^2+p^2\right)=0\) Therefore, Locus of \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is \(2 a x+2 b y-\left(a^2+b^2+p^2\right)=0\)
